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ELEMENTS 
OF PLANE AND SPHERICAL 

TRIGONOMETRY, 

WITH THEIR APPLICATIONS TO 

MENSURATION, SURYEYING, AND NAYIGATION. 



BY 



ELIAS LOOMIS, LL.D., 

PROFESSOR OF XATURAL PHILOSOPHY AND ASTR0X03IT IX TALE COLLEGE, 
AUTHOR OF A "COURSE OF MATHEMATICS," ETC. 



,1 ■ ' 



REVISED EDITION. 




NEW YORK: 

HARPER & BROTHERS, FRANKLIN SQUARE. 

1886. 






K-5'^ 



LOOMIS'S SERIES OF TEXT-BOOKS. 



ELEMENTARY ARITHMETIC. 165 pp., 28 cents, 

ELEMENTS OF ALGEBRA. Revised Edition, 281 pp., 90 cents. 

Key to Elements of Algebra, for Use of Teachers, 128 pp., 90 cents. 
TREATISE ON ALGEBRA. Revised Edition. 384 pp., $1 00. 

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ELEMENTS OF GEOMETRY. Revised Edition. 388 pp., $1 00. 
ELEMENTS OF TRIGONOMETRY, SURVEYING, AND NAVIGATION. Revised Edi- 

tion. 194 pp., $1 00. 
TABLES OF LOGARITHMS. ISO pp., $1 00. 

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Published by HARPER & BROTHERS, New York. 
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Entered, according to Act of Congress, in the year one thousand 
eight hundred and fifty-eight, by 

Harper & Brothers, 

In the Clerk's Office of the District Court of the Southern District 
of New York. 



Copyright, 1S86, by Harper & Brothers. 



^ I ^ 



PREFACE. 



The stereotype plates of my " Elements of Trigonometry " 
liaving become so much worn in the printing of more than 
60,000 copies that they were no longer fit for use, it became 
necessary to recast them, and I have improved the opportunity 
to make a thorough revision of the work. At the same time 
I have introduced a radical change which I have had in con- 
templation for several years. In the former editions, in con- 
formity with the usage of the old English mathematicians, 
the trigonometrical functions were regarded as lines ; in this 
revised edition they are regarded as ratios, in conformity with 
a usage which has now become well-nigh universal. I am of 
the opinion that the old system has some important advantages 
in the training of students who have no decided aptitude 
for mathematical studies ; but since the weight of authority 
is decidedly against this system, I have decided to abandon it. 
It is hoped that the changes made in this revised edition may 
receive the general approval of teachers. 



CONTENTS. 



CHAPTER I. 

THE NATURE AND PROPERTIES OF LOGARITHMS. 

Page 

Nature of Logarithms 7 

Description of the Table of Logarithms 9 

Multiplication by Logarithms 14 

Division by Logarithms 16 

Involution by Logarithms 17 

Evolution by Logarithms 18 

Proportion by Logarithms 19 

CHAPTER H. 

PLANE TRIGONOMETRY. 

Sines, Tangents, Secants, &c. , defined 22 

Explanation of the Trigonometrical Tables 26 

To find Sines and Tangents of Small Arcs 32 

Solution of Right-angled Triangles 35 

Solution of Oblique-angled Triangles 39 

Instruments used in Drawing 45 

Geometrical Construction of Triangles 49 

Values of the Sines, Cosines, &c., of certain Angles. 51 

Trigonometrical Formulse 55 

Computation of a Table of Sines, Cosines, &c 59 

Examples for Practice 62 

CHAPTER IH. 

MENSURATION OF SURFACES AND SOLIDS. 

Areas of Figures bounded by Right Lines 66 

Area of a Regular Polygon 71 

Quadrature of the Circle and its Parts 75 

Mensuration of Solids 78 

Railway Excavations or Embankments 83 

Regular Polyedrons 88 

The Three Round Bodies 90 

Area of a Spherical Triangle , 94 

Examples for Practice 95 



vi Contents. 

CHAPTER IV. 

SURVEYING. 

Page 

Definitions 100 

Instruments for Measuring Angles 101 

Explanation of the Yernier 104 

Description of tlie Theodolite „ 105 

Heights and Distances 107 

The Determination of Areas „ 113 

Plotting a Survey _ . , 114 

The Traverse Table 116 

To Find the Area of a Field 119 

Trigonometrical Surveys 124 

Variation of the Needle , 127 

Levelling 130 

Topographical Maps ,.,.., 134 

Setting out Railway Curves 138 

Surveying Harbors ,..,..., 141 

The Plane Table i42 

To Determine the Depth of Water i44 

Examples for Practice 145 

CHAPTER Y. 

NAVIGATION. 

Definitions, &c 150 

Plane Sailing 153 

Traverse Sailing , 155 

Parallel Sailing „ I59 

Middle Latitude Sailing 161 

Mercator's Sailing 164 

Nautical Charts 168 

Examples for Practice 169 

CHAPTER VI. 

SPHERICAL TRIGONOMETRY. 

Right-angled Spherical Triangles , I73 

Napier's Rule of the Circular Parts 176 

Examples of Right-angled Triangles 179 

Oblique-angled Spherical Triangles 181 

Examples of Oblique-angled Triangles 183 

Trigonometrical Formulae 189 

Sailing on an Arc of a Great Circle , , 194 

Examples for Practice ,. = ..,.,.... 106 



TRIGONOMETRY. 



CHAPTER I. 

THE NATURE AND PROPERTIES OF LOGARITHMS. 

Article 1. Logarithms are numbers designed to diminish 
thelaborof Multiplication and Division, by substituting in their 
stead Addition and Subtraction. All numbers are res^arded as 
powers of some one number, which is called the lase of the 
system ; and the exponent of that power of the base which is 
equal to a given number is called the logarithm of that number. 

The base of the common system of logarithms (called, from 
their inventor, Briggs's logarithms) is the number 10. Hence 
all numbers are to be regarded as powers of 10. Thus, since 
W — 1^ is the logarithm of 1 in Briggs's system; 

10^ = 10, 1 '' " 10 ^'^ " 

10^ = 100, 2 " " 100 " " 

10^ = 1000, 3 " " 1000 " " 

10^ =.10,000, 4 " " 10,000 " " 

&c., &c., &c. ; 

whence it appears that, in Briggs's system, the logarithm of 
every number between 1 and 10 is some number between 
and 1, i. 6., is a proper fraction. The logarithm of every num- 
ber between 10 and 100 is some number between 1 and 2, i. e., 
is 1 plus a fraction. The logarithm of every number between 
100 and 1000 is some number between 2 and 3, i. e., is 2 plus a 
fraction, and so on. 

2. The preceding principles may be extended to fractions 
by means of negative exponents. Thus, since 
10-^ = 0.1, —1 is the logarithm of 0.1 in Briggs's system ; 
10-^^=0.01, -2 " " 0.01 " " 

10-3=0.001, -3 " " 0.001 " " 

10-'' = 0.0001,-4 " " 0.0001 " " 

&c., &c., &c. 



8 Teigonometey. 

Hence it appears that the logarithm of every number between 
1 and 0.1 is some number between and —1, or may be rep- 
resented by —1 plus a fraction ; the logarithm of every num- 
ber between 0.1 and .01 is some number between —1 and —2, 
or may be represented by —2 plus a fraction ; the logarithm 
of every number between .01 and .001 is some number be- 
tween — 2 and —3, or is equal to —3 plus a fraction, and 
80 on. 

The logarithms of most numbers, therefore, consist of an 
integer and a fraction. The integral part of a logarithm is 
called its characteristic, and the decimal part is called its 7nan- 
iissa. The characteristic may be known from the following 

EULE. 

The characteristic of the logarithm of any number greater 
than unity, is one less than the number of integral figures in 
the given number. 

Thus the logarithm of 297 is 2 plus a fraction ; that is, the 
characteristic of the logarithm of 297 is 2, wdiich is one less 
than the number of integral figures. The characteristic of the 
logarithm of 5673.29 is 3 ; that of 73254.1 is 4, etc. 

The characteristic of the logarithm of a decimal fraction is 
a negative number, and is equal to the number of f laces by which 
its first significant figure is removed from the place of units. 

Thus the logarithm of .0€46 is ~3 plus a fraction ; that is, 
the characteristic of the logarithm is —3, the first significant 
figure, 4, being removed three places from units. 

3. Since powers of the same quantity are multiplied by 
adding their exponents {Alg., Art. 58), 

The logarithm of the product of two or inore factors is equal 
to the sum of the logarithms of those factors. 

Hence we see that if it is required to multiply two or more 
numbers by each other, we have only to add their logarithms : 
the sum will be the logarithm of their product. We then 
look in the table for the number answering to that logarithm, 
in order to obtain the required product. 

Also, since powers of the same quantity are divided by sub- 
tracting their exponents {Alg., Art. 72), 

The logarithm of the quotient of one number divided by 



Logarithms. 9 

another^ is equal to the difference of the logarithms of those 
nitmbers. 

Hence we see that if we wish to divide one number by an- 
other, we have only to subtract the logarithm of the divisor 
from that of the dividend ; the difference will be the logarithm 
of their quotient. 

4. Since, in Briggs's system, the logarithm of 10 is 1, if 
any number be multiplied or divided by 10, its logarithm will 
be increased or diminished by 1 ; and as this is an integer, it 
will only change the characteristic of the logarithm, without 
affecting the decimal part. Hence 

The decimal part of the logarithm of any number is the 
same as that of the number midtiplied or divided by 10, 100, 
1000, cfcc. 

Thus, the logarithm of 65430 is 4.815777; 

" " 6543 is 3.815777; 



" " 654.3 is 2.81577 



il u 



65.43 is 1.815777; 
" " ^ 6.543 is 0.815777; 

« " .6543 is 1.815777; 

.06543 is 2.815777; 
.006543 is 3.815777. 



a a 



The minus sign is here placed over the characteristic, to 
show that that alone is negative, w^iile the decimal part of the 
logarithm is positive. 

Table of Logarithms. 

5. A table of logarithms usually contains the logarithms 
of the entire series of natural numbers from 1 up to 10,000, 
and the larger tables extend to 100,000 or more. In the smaller 
tables, the logarithms are usually given to five or six decimal 
places ; the larger tables extend to seven, and sometimes eight 
or more places. 

In the accompanying table, the logarithms of the first 100 
numbers are given with their characteristics; but, for all oth- 
er numbers, the decimal part only of the logarithm is given, 
while the characteristic is left to be supplied, according to the 
rule in Art. 2. 



10 Tkigonometry. 

6. To find the Logarithm of any Numher between 1 and 100. 

Look on the first page of the accompanying table, along 

the column of numbers under ]^., for the given number, and 

against it, in the next column, will be found the logarithm 

with its characteristic. Thus, 

opposite 13 is 1.113943, which is the logarithm of 13; 
" 65 is 1.812913, " '' 65. 

To find the Logarithm of any Numher consisting of Three 

Figures. 

Look on one of the pages of the table from 2 to 20, along 
the left-hand column, marked E"., for the given number, and 
against it, in the column headed 0, will be found the decimal 
part of its logarithm. To this the characteristic must be pre- 
fixed, according to the rule in Art. 2.* Thus 
the logarithm of 347 will be found, from page 8, 2.540329 ; 

'' " 871 " " 18, 2.940018. 

As the first two figures of the decimal are the same for sev- 
eral successive numbers in the table, they are not repeated for 
each logarithm separately, but are left to be supplied. Thus 
the decimal part of the logarithm of 339 is .530200. The first 
two figures of the decimal remain the same up to 347; they 
are therefore omitted in the table, and are to be supplied. 

To find the Logarithm of any Numher consisting of Foicr 

Figures. 

Find the three left-hand figures in the column marked N., 
as before, and the fourth figure at the head of one of the oth- 
er columns. Opposite to the first three figures, and in the col- 
umn under the fourth figure, will be found four figures of 
the logarithm, to which two figures from the column headed 
are to be prefixed, as in the former case. The characteristic 
must be supplied according to Art. 2. Thus 

the logarithm of 3456 is 3.538574; 
'' " 8765 is 3.942752. 

In several of the columns headed 1, 2, 3, &c., small dots are 
found in the place of figures. This is to show that the two 
figures which are to be prefixed from the first column have 
changed, and they are to be taken from the horizontal line di- 



LOGAEITHMS. 11 

rectly lelow. The place of the dots is to be supplied with ci- 
phers. Thus 

the logarithm of 2045 is 3.310693 ; 

" " 9777 is 3.990206. 

The two leading figures from the column must also be 

taken from the horizontal line below, if any dots have been 

passed over on the same horizontal line. Thus 

the logarithm of 1628 is 3.211654 

To find the Logarithm of any Nicmber containing more than 
Four Figures. 

7. By inspecting the table, we shall find that, within cer- 
tain limits, the differences of the logarithms are nearly propor- 
tional to the differences of their corresponding numbers. Thus 
the logarithm of 7250 is 3.860338 ; 
" ^ " 7251 is 3.860398; 

" " 7252 is 3.860458 ; 

" " 7253 is 3.860518. 

Here the difference between the successive logarithms, called 
the tabular difference, is constantly 60, corresponding to a dif- 
ference of unity in the natural numbers. If, then, we suppose 
the differences of the logarithms to be proportional to the dif- 
ferences of their corresponding numbers (as they are nearly), 
a difference of 0.1 in the numbers should correspond to a 
difference of 6 in the logarithms; a difference of 0.2 in the 
numbers should correspond to a difference of 12 in the log- 
arithms, &c. Hence 

the logarithm of 7250.1 must be 3.860344; 
" " 7250.2 " 3.860350; 

" " 7250.3 " 3.860356. 

In order to facilitate the computation, the tabular difference 
is inserted on page 16 in the column headed D., and the pro- 
portional part for the fifth figure of the natural number is 
given at the bottom of the page. Thus, when the tabular 
difference is 60, the corrections for .1, .2, .3, &e., are seen to 
be 6, 12, 18, etc. 

If the given number was 72501, the characteristic of its log- 
arithm would be 4, but the decimal part would be the same as 
for 7250.1. 



12 Tkigonometey. 

If it is required to find the correction for a sixth figure 
in the natural number, it is readily obtained from the Propor- 
tional Parts in the table. The correction for a figure in the 
sixth place must be one tenth of the correction for the same 
figure if it stood in the fifth place. Thus, if the correction for 
.5 is 30, the correction for .05 is obviously 3. 

As the differences change rapidly in the first part of the 
table, it was found inconvenient to give the proportional parts 
for each tabular difference ; accordingly, for the first seven 
pages, they are only given for the even differences, but the 
proportional parts for the odd differences will be readily found 
by inspection. 

Eequired the logarithm of 452789. 

The logarithm of 452700 is 5.655810. 

The tabular difference is 96. 

Accordingly, the correction for the fifth figure, 8, is 77, and 
for the sixth figure, 9, is 8.6, or 9 nearly. Adding these cor- 
rections to the number before found, we obtain 5.655896. 

It is not claimed that tlie preceding logarithms are perfect- 
ly correct, but simply that they are the nearest values limited 
to six decimal places. Accordingly, when the fraction which 
is omitted exceeds half a unit in the sixth decimal place, the 
last figure must be increased by unity. 

Eequired the logarithm of 8765432. 

The logarithm of 8765000 is 6.942752 

Correction for the fifth figure, 4, 19.6 

" " sixth figure, 3, 1.5 

" " seventh figure, 2, 0.1 

Therefore the logarithm of 8765432 is 6.942773 

Eequired the logarithm of 234567. 

The logarithm of 234500 is 5.370143 

Correction for the fifth figure, 6, 111 

" " sixth figure, 7, 13 

Therefore the logarithm of 234567 is 5.370267 

To find the Logarithm of a Decimal Fraction. 

8. According to Art. 4, the decimal part of the logarithm 
of any number is the same as that of the number multiplied 



LOGAEITHMS. 13 

or divided by 10, 100, 1000, &c. Hence, for a decimal frac- 
tion, we find the logarithm as if the figures were integers, and 
prefix the characteristic according to the rule of Art. 2. 





Examples. 




The logarithm of 3i5.6 


is 2.538571:; 


a a 


87.65 


is 1.942752; 


u a 


2M5 


is 0.370143; 


a a 


.1234 


is 1.091315; 


u a 


OOf^fiY! 


^ IS ^ ^KztlQn 



To find the Logarithm of a Vulgar Fraction. 

9. We may reduce the vulgar fraction to a decimal, and 
find its logarithm by the preceding article; or, since the value 
of a fraction is equal to the quotient of the numerator divided 
by the denominator, we may, according to Art. 3, subtract the 
logarithm of ths denominator from that of the numerator ; 
the difference will be the logarithm of tlLe fraction. 

Ex. 1. Find the logarithm of -f^, or 0.1875. 

From the logarithm of 3, 0.477121 

Take the logarithm of 16, 1.204120 

Leaves the logarithm of A? or -IS <^Sj 1.273001 
Ex. 2. The logarithm of -/^ is 2.861697. 
Ex. 3. The logarithm of ^ is 1.147401. 

To find the Natural Niunler corresjponding to any Logarithm.. 

10. Look in the table, in the column headed 0, for the first 
two figures of the logarithm, neglecting the characteristic ; 
the other four figures are to be looked for in the same col- 
umn, or in one of the nine following columns ; and if they 
are exactly found, the first three figures of the corresponding 
number will be found opposite to them in the column headed 
N., and the fourth figure will be found at the top of the page. 
This number must be made to correspond with the charac- 
teristic of the given logarithm, by pointing off decimals or 
annexing ciphers. Thus 

the natural mimber belonging to the log. 4.370143 is 23450; 
« '' " " 1.538574 is 34.56. 



14: Teigonometey. 

If the decimal part of the logarithm cannot be exactly found 
in the table, look for the nearest less logarithm, and take out 
the four figures of the corresponding natural number as be- 
fore; the additional figures may be obtained by means of the 
Proportional Parts at the bottom of the page. 

Required the number belonging to the logarithm 4.368399. 

On page 6, we find the next less logarithm .368287. 

The four corresponding figures of the natural number are 
2335. Their logarithm is less than the one proposed by 112. 
The tabular difference is 186; and, by referring to the bottom 
of page 6, we find that, with a difference of 186, the figure 
corresponding to the proportional part 112 is 6. Hence the 
Rve figures of the natural number are 23356 ; and, since the 
characteristic of the proposed logarithm is 4, these five figures 
are all integral. 

Required the number belonging to logarithm 5.345678. 

The next less logarithm in the table is 345570. 

Their difference is 108. 

The first four figures of the natural number are 2216. 

With the tabular difference 196, the fifth figure, correspond- 
ing to 108, is seen to be 5, with a remainder of 10. To find 
the sixth figure corresponding to this remainder 10, we may 
multiply it by 10, making 100, and search for 100 in the 
same line of proportional parts. We see that a difference 
of 100 would give us 5 in the fifth place of the natural num- 
ber. 'Therefore, a difference of 10 must give us 5 in the 
sixth place of the natural number. Hence the required num- 
ber is 221655. 

In the same manner we find 

the number corresponding to log. 3.538672 is 3456.78 ; 
" " " 1.994605 is 98.7654; 

" " " 1.647817 is .444444. 



Multiplication by Logaeithms. 

11. According to Art. 3, the logarithm of the product of 
two or more factors is equal to the sum of the logarithms of 
those factors. Hence, for multiplication by Jpgarithms, we 
have the following 



Logarithms. 15 

EULE. 

Add the logarithms of the factors ; the sum vnll le the log- 
arithm of their product. 

Ex. 1. Eequired the product of 57.98 bj 18. 

The logarithm of 57.98 is 1.763278 

'' " 18 is 1.255273 



The logarithm of the product 1043.64 is 3.018551 
Ex. 2. Keqiiired the product of 397.65 by 43.78. 

Ans., 17409.117. 
Ex. 3. Eequired the continued product of 54.32, 6.543, and 
12.345. 

The word sum^in the preceding rule, is to be understood in 
its algebraic sense ; therefore, if any of the characteristics of 
the logarithms are negative, we must take the difference be- 
tween their sum and that of the positive characteristics, and 
prefix the sign of the greater. It should be remembered that 
the decimal part of the logarithm is invariably positive ; hence 
that which is carried from the decimal part to the character- 
istic must be considered positive. 
Ex. 4. Multiply 0.00563 by 17. 

The logarithm of 0.00563 is 3.750508 
" " 17 is 1.230449 



Product, 0.09571, whose logarithm is 2.980957 

Ex. 5. Multiply 0.3854 by 0.0576. Ans., 0.022199. 

Ex. 6. Multiply 0.007853 by 0.00476. 

Ans., 0.00003738. 
Ex. 7. Find the continued product of 11.35, 0.072, and 0.017. 
12. Negative quantities may be multiplied by means of 
logarithms in the same manner as positive, the proper sign 
being prefixed to the result according to the rales of Algebra. 
To distinguish the negative sign of a natural number from 
the negative characteristic of a logarithm, we append the let- 
ter n to the logarithm of a negative factor. Thus 
the logarithm of —56 we write 1.748188 n. 
Ex. 8. Multiply 53.46 by -29.47. 

The logarithm of 53.46 is 1.728029 
" " -29.47 is 1.469380 n. 

Product, -1575.47, log. 3.197409 n. 



16 Teigonometkt. 

Ex. 9. Find the continued product of 3Y2.1, —.0054, and 
-175.6. 

Ex. 10. Find the continued product of -0.137, -7.689. 
iind -.0376. 

Division by Logaeithms. 

13. According to Art. 3, the logarithm of the quotient of 
one number divided by another is found by subtracting the 
logarithm of the divisor from that of the dividend. Hence, 
for division by logarithms, we have the following 

KULE. 

From the logarithm of the dividend, subtract the logarithm 
of the divisor ; the difference will he the logarithm of the 
quotient, 

Ex. 1. Required the quotient of 888.7 divided by 42.24. 
The logarithm of 888.7 is 2.948755 
" 42.24 is 1.625724 

The quotient is 21.039, whose log. is 1.323031 
Ex. 2. Required the quotient of 3807.6 divided by 13.7. 

Ans., 277.927. 
The word difference, in the preceding rule, is to be un- 
derstood in its algebraic sense ; therefore, if the character- 
istic of one of the logarithms is negative, or the lower one 
is greater than tlie upper, we must change the sign of the 
subtrahend, and proceed as in addition. If unity is carried 
from the decimal part, this must be considered as positive, 
and must be united with the characteristic before its sign is 
changed. 

Ex. 3. Required the quotient of 56.4 divided by 0.00015. 
The logarithm of 56.4 is 1.751279 
" '^ 0.00015 is 4.176091 

The quotient is 376000, whose log. is 5.575188 
This result may be verified in the same way as subtraction 
in common arithmetic. The remainder, added to the subtra- 
hend, should be equal to the minuend. This precaution 
should always be observed when there is any doubt with re- 
gard to tlie sign of the result. 



Logarithms. 17 

Ex. 4. Kequired the quotient of .8692 divided bv 42.258. 

Ans. 

Ex. 5. Eequired the quotient of .74274 divided by .00928. 

Ex. 6. Eequired the quotient of 24.934 divided by .078541. 

Negative quantities may be divided by means of logarithms 
in the same manner as positive, the proper sign being pre- 
fixed to the result according to the rules of xVlgebra. 

Ex. 7. Eequired the quotient of -79.54 divided by 0.08321. 

Ex. 8. Eequired the quotient of -0.4753 divided by —36.74. 

IxTOLrTION BY LOGARITHMS. 

14. It is proved in Algebra, Art. 398, that the logarithm 
of any power of a number is equal to the logarithm of that 
number multiplied by the exponent of the power. Hence, to 
involve a number by logarithms, we have the following 

EULE. 

Multiply the logarithm of the nwiiber hj the exjponent of 
the power required. 

Ex. 1. Eequired the square of 428. 

The logarithm of 428 is 2.631444 

2 

Square, 183184, log. 5.262888 
Ex. 2. Eequired the 20th power of 1.06. 

The logarithm of 1.06 is 0.025306 

20 

20th power, 3.2071, log. 0.506120 
Ex. 3. Eequired the 5th power of 2.846. 
It should be remembered, that what is carried from the dec- 
imal part of the logarithm is positive, whether the characteris- 
tic is positive or negative. 

Ex. 4. Eequired the cube of .07654. 

The logarithm of .07654 is 2.883888 

3 

Cube, .0004484, log. 4.651664 
Ex. 5. Eequired the fourth power of 0.09874. 
Ex. 6. Eequired the seventh power of 0.8952. 

B 



18 Tkigonometkt. 

Evolution by Logarithms . 
15. It is proved in Algebra, Art. 399, that the logarithm 
of any root of a number is equal to the logarithm of that num- 
ber divided by the index of the root. Hence, to extract the 
root of a number by logarithms, we have the following 

Rule. 

Divide the logarithm of the miinber hy the index of the root 
Teqiiired, 

Ex. 1. Eequired the cube root of 482.38. 

The logarithm of 482.38 is 2.683389. 

Dividing by 3, we have 0.894463, which corresponds to 
7.8426, which is therefore the root required. 

Ex. 2. Required the 100th root of 365. 

Ans., 1.0608. 

When the characteristic of the logarithm is negative, and is 
not divisible by the given divisor, we may increase the char- 
acteristic by any number which will make it exactly divisible, 
provided we prefix an equal positive number to the decimal 
part of the logarithm. 

Ex. 3. Required the seventh root of 0.005846. 

The logarithm of 0.005846 is 3.766859, which may be writ- 
ten 7+4.766859. 

Dividing by 7, we have 1.680980, which is the logarithm of 
.4797, which is, therefore, the root required. 

This result may be verified by multiplying l."6S0980 by 7; 
the result will be found to be 3.766860. 

Ex. 4. Required the fifth root of 0.08452. 

Ex. 5. Required the tenth root of 0.007815. 

16. By combining the principles of Arts. 14 and 15, we 
may find any power of any root of a given quantity. 

Ex. 1. Required tlie cube of the seventh root (or the seventh 
root of the cube) of 52.87. 

The logarithm of 52.87 is 1.723209 

multiplying the logarithm by 3 3 

we obtain 5.169627 

Dividing the product by 7, we obtain 0.738518 
which is the logarithm of 5.4767, Ans. 



LOGAEITHMS. 19 

Ex. 2. Kequired the tenth power of the uintli root of 0.7329. 

A71S., 0.70803. 
Ex. 3. Eequired the fifth root of the third power of 0.02718. 

Ans. 
Ex. 4. Kequired the sixth root of the fifth power of 0.3829. 

Ans. 

Pkopoetion by Logaeithms. 

17. The fourth term of a proportion is found bj multi- 
plying together the second and third terms, and dividing by 
the first. Hence, to find the fourth term of a proportion by 
logarithms, 

Add the logarithms of the second and third terms ^ and from 
their sum subtract the logarithm of the first term. 

Ex. 1. Find a fourth proportional to 72.34, 2.519, and 
357.48. Ans,, 12.448. 

Ex. 2. Find a fourth proportional to 26.71, 412.3, and 
0.07934. Ans. 

Ex. 3. Find a fourth proportional to —25.93, 4.037, and 
0.52783. Ans. 

18. When one logarithm is to be subtracted from another, 
it may be more convenient to convert the subtraction into an 
addition, which may be done by first subtracting the given 
logarithm from 10, adding the difierence to the other log- 
arithm, and afterward rejecting the 10. 

TJie difference between a given logarithm and 10 is called 
its arithmetical comjplement., or eologaritlim j and this is easily 
found from the table by beginning with the characteristic, and 
subtracting each successive figure of the logarithm from 9, 
except the last significant figure on the right, which must be 
subtracted from 10. 

To subtract one logarithm from another is the same as to 
add its complement, and then reject 10 from the result. For 
a—h'i^ equivalent to 10 — Z> + a — 10. 

To work a proportion, then, by logarithms, we must 

Add the com/plement of the logarithm of the first term to the 
logarithms of the second and third terms. 

The characteristic must afterward be diminished by 10. 

Ex. 1. Find a fourth proportional to 6853, 489, and 38750. 



20 Teigonometry. 

The complement of the logarithm of 6853 is 6.164119 
The logarithm of 489 is 2.689309 

38750 is 4.588272 
The fourth term is 2765, whose logarithm is 3.441700 
One advantage of using the complement of the first term 
in working a proportion by logarithms is, that it enables us to 
exhibit the operation in a more compact form. 

Ex. 2. Find a fourth proportional to 73.84, 658.3, and 4872. 

A71S. 
Ex. 3. Find a fourth proportional to 5.745, 781.2, and 54.27. 



CHAPTEE II. 

PLANE TRIGOXOMETRY. 

19. Trigonometry is the science which treats of the relations 
of the sides and angles of a triangle ; and also embraces inves- 
tigations respecting the relations of angles in general. 

Plane Trigonometry treats of triangles on a plane surface ; 
and Spherical Trigonometry treats of triangles on a spherical 
surface. 

20. Each triangle has six parts ; three angles and three 
sides. AYhen any three of these parts are given, provided one 
of them is a side, the other parts may be determined. By the 
solution of a triangle is meant the determination of the un- 
known parts of a triangle when certain parts are given. 

21. The magnitude of an angle is expressed by degrees^ 
minutes^ and seconds ; distinguished by the characters ° ' " . 

A degree is an angle equal to -^ of a right angle. A right 
angle is therefore expressed by 90° ; two right angles by 
180° ; and four right angles (or the whole angular space about 
a point) by 360°. 

A minute is an angle equal to -^-^ of a degree. Hence 1° = 
60'; and a right angle = 90 x 60' ==5400'. 

A second is an angle equal to -g^ of a minute. Hence V — 
60"; 1°=^ 60x60" = 3600"; and a right angle = 90 x 60x60" 
= 324000". 

22. Since the angles at the centre of a circle are propor- 
tional to the arcs of the circumference intercepted between 
their sides, these arcs may be taken as the measures of the in- 
cluded angles; and the arcs may be expressed by degrees, 
minutes, and seconds. A degree of arc is the arc which sub- 
tends an angle of one degree at the centre of the circle ; that 
is, it is 3-|-g- of the circumference of the circle. A quadrant is 
expressed by 90°; a semi-circumference by 180°; and an en- 
tire circumference by 360°. 



22 



Trigonometry. 



23. The complement of an angle or arc is what remains 
after subtracting the angle or arc from 90°. Thus the com- 
plement of 30° is 60° ; and the complement of 25° 15^ is 
64° 45^ 

In general, if we represent any arc by A, its complement is 
90° — A. Hence the complement of an arc that exceeds 90° 
is negative. Thus the complement of 120° is 90° -120° = 
-30°. The complement of 100° 15' is -10° 15^ 

Since the two acute angles of a right-angled triangle are to- 
gether equal to a right angle, each of them is the complement 
of the other. 

24. The supjplement of an angle or arc is what remains after 
subtracting the angle or arc from 180°. Thus the supplement 
of 30° is 150° ; and the supplement of 25° 15' is 154° 45'. 

In general, if we represent any arc by A, its supplement is 
180°— A. Hence the supplement of an arc that exceeds 180° 
is negative. Thus the supplement of 200° is 180° -200° = 
-20°. 

Since, in every triangle, the sum of the three angles is 
180°, either angle is the supplement of the sum of the other 
two. 

Let ABC, ADE, AFG, &c., be a series 
of right-angled triangles which have a 
common angle A. Since the angles at 
C, E, and G are equal to each other, the 




triangles are similar, and w^e have 



BC 



AB::DE: AD::FG: AF; 

BC_DE_FG 

AB~AD"AF' 

BC _ DE _ FG 

AC~AE~AG' 

AB_AD_ AF. 

AC~AE~"AG' 
that is, in all onght- angled triangles having the same acute 
angle, the sides have to each other the same ratio. These ratios 
have received special names, as follows : 

26. The sine of the angle A is the quotient of the opposite 
side divided hy the hypothenuse. 



that is, 



Also = 



and 



Plane Teigonometkt. 



23 




3 Thus in the right-angled triangle ABC, 
if we designate the sides by the small let- 
ters a^ 5, <?, we shall have 

Sine 01 A = — ; sine or 13 ::= — . 

G G 

27. The tangent of the angle A is the quotient of the op- 
posite side divided hy the adjacent side. 

Thus, 



tano^ent of A = -y-; tano^ent of B = — 



28. The secant of the angle A is the quotient of the hypoth- 
enuse divided hy the adjacent side. 



Thus, secant of A 



secant of B = — . 



b ' a 

29. The cosine, cotangent^ and cosecant of an angle are re- 
sjyectively the sine, tangent, and secant of the complement of 
that angle. 

Since the two acute angles of a right-angled triangle are 
complements of each other, we shall have 

a 



sine of A = cosine of B = 

tan. of A = cotan. of B = 

secant of A = cosec. of B — 



cosine of A = sine of B = 



cotan. of A = tan. of B=: 



c ' 
a 

o 

7- : cosec. of A = secant of B =: — . 
b a 

The words sine, cosine, &c., are usually abbreviated thus : 
for sine of A we write sin. A; for cosine of A we write 
cos. A, etc. 

The quantities sine, cosine, tangent, &c., are sometimes des- 
ignated by the term trigonometric functions. 

30. Since the angles at the centre of a circle are propor- 
tional to the arcs of the circumference intercepted between 
their sides, the sine, tangent, and secant of an angle may be 
regarded as the sine, tangent, and secant of the arc which 
D measures that angle. 

Thus, the sine of the angle BAC = the 

sine of the arc BE = . 

AB 

31. If the radius of the circle be taken 




24: 



Teigonometky. 



equal to unity, the trigonometric functions above defined may 
be represented by straight lines. 

Representing the angle BAG, or the arc BE by Aj and sup- 
posing AB or AEi=l, we have 

BCBC 

AB~ 

DE 

ae' 

AD 



sin. A 



tan. A — 



= DE 



sec. A — 



AD. 



^ -BC; 

DE 
1 
_AD 
AE ~ 1 

Therefore in a circle whose radius is unity, the sine of an 
arc, or of the angle at the centre ineasured hy that arc, is the 
perpendicular let fall from one extremity of the arc ^ipon the 
diameter passing through the other extremity. 

The tangent of an arc is the line lohich touches the circle at 
one extremity of the arc, and is limited hy the diameter {pro- 
duced) passing through the other extremity. 

The secant of an arc is that part of the ijroduced diameter 
lohich is intercepted hetioeen the centre and the tangent. 

In a circle whose radius is not supposed to be unity, the 
trigonometric functions of an arc w^ill be equal to the lines 
here defined, divided by the radius of the circle. 

The properties here enumerated may be employed as the 
I definitions of the terms sine, tangent, 
and secant. The annexed figure il- 
lustrates these definitions, and will 
be referred to hereafter. 

32. The cosine of- an arc is the 
sine of the complement of that arc. 
Thus since the arc DE is the com- 
plement of AF, FK is the sine of the 
arc DE or the cosine of the arc AF. 
The cotangent of an arc is the tan- 
gent of the complement of that arc. Thus DL is the tangent 
of the arc DF, or the cotangent of the arc AF. 

The cosecant of an arc is the secant of the complement of 
that arc. Thus CL is the secant of the arc DF, or the cose- 
cant of the arc AF. 




Plane Tkigonometey. 25 

The versed sine of an arc is that part of tlie diameter inter- 
cepted between the extremity of the arc, and the foot of the 
sine. Thus GA is the versed sine of the arc AF. 

33. The sine of an arc is equal to half the chord of twice 
the arc, the radius of the circle being supposed unity. Thus 
the sine FG is the half of FH, which is the cliord of the arc 
FAH, which is double of FA. 

34. The sine of the supplement of an arc is the same as 
thai of the arc itself Thus FG is the sine of the arc AF, or 
of its supplement BDF. 

The tangent of the arc BDF is BM, wdiich is numerically 
equal to AI, the tangent of the arc AF; but being measured 
in the opposite direction from the line AB, we attribute to it 
the opposite algebraic sign. 

The secant of the arc BDF is CM, which is numerically 
equal to CI, the secant of the arc AF. Thus we see that the 
sine, tangent, and secant of an arc, are numerically equal to 
the sine, tangent, and secant of its supplement ; but it does 
not hence follow that they should have the same algebraic 
sign. This question will be considered in Art. 79. 

35. The relations of sines, cosines, &c., to each other may 
be derived from Arts. 26-29, or from the properties of simi- 
lar triangles. Thus, since the triangles CGF, CAT, and CDL 
are similar, \\q have 

1. CG : GF : : CA : AI ; that is, representing the arc by A, 
and assuming the radius of the circle to be unity, we have 
COS. A : sin. A : : 1 : tan. A. 

Wiience tan. A = — '- — . 









COS. A 




2. 


GF: 


CG: 


: CD : DL ; that is, sin. A : cos. 

Whence cotan. A = --- — . 
sin. A 


A : : 1 : cotan. A. 


3. 


CG 


:CF: 


: CA : CI ; that is, cos. A : 1 : 


: 1 : sec. A. 




cos. A 




4. 


GF: 


:CF: 


: CD : CL ; that is, sin. A : 1 

Whence cosec. A = . 

sin. A 


: : 1 : cosec. A. 



26 



Teigonometky. 



5. AI : AC ; : CD : BL ; that is, tan. A : 1 : : 1 : cotan. A. 

Whence tan. A = . 

cot. A 

36. In the ri^ht-angled triangle CGF we have 

Gr^ + CG^=CF = 1 ; that is, sin. ^A + cos. ^A:=l. 

in which the notation sin. ^A signifies "the square of the sine 

of A." Hence the square of the sine of an arc, together with 

the square of its cosine, is equal to unity. 

From this formula we find 

sin. A=: + i/l — cos. ^A; and cos. A^Hh-v/l — sin.^ A. 

37. A table of natural sines, tangents, &c., is a table giv- 
ing the lengths of these lines for different angles in a circle 
whose radius is unity. 

Thus, if we describe a circle with a radius of one inch, and 
divide the circumference into equal parts po» 
of ten degrees, we shall find 

the sine of 10° equals 0.174 inch ; 







20° 




0.342 








30° 




0.500 








40° 




0.643 








50° 




0.766 








60° 




0.866 








70° 




0.940 








80° 




0.9S5 








90° 




1.000 






If we draw the tangents of the same 
arcs, we shall find 

the tangent of 10° equals 0.176 inch ; 





u 


20° 




0.364 " 




a 


30° 




0.577 " 




u 


40° 




0.839 " 




u 


45° 




1.000 " 




a 


50° 




1.192 '^ 




a 


60° 




1.732 " 




u 


70° 




2.747 " 




u 


80° 




5.671 " 




a 


90° 




infinite. 



Also, if we draw the secants of the same 
arcs, we shall find that 




Plane Teigonometkt. 2:7 

the secant of 10° equals 1.015 inch; 



20° 


a 


1.064: ' 


30° 


a 


1.155 ' 


40° 


a 


1.305 ' 


50° 


u 


1.556 ^ 


60° 


a 


2.000 < 


70° 


a 


2.921: ' 


80° 


a 


5.759 ' 


90° 


u 


infinite. 



In the accompanying table, pages 116-133, the sines, co- 
sines, tangents, and cotangents are given for every minute of 
the quadrant to six places of figures. 

38. To find from the tcible the natural sine, cosine, c&c, of 
an arc or angle. 

If a sine is required, look for the degrees at the top of the 
page, and for the minutes on the left ; then, directly under the 
given number of degrees at the top of the page, and opposite 
to the minutes on the left, will be found the sine required. 
Since the radius of the circle is supposed to be unity, the sine 
of every arc below 90° is less than unity. The sines are ex- 
pressed in decimal parts of radius ; and, although the decimal 
point is not written in the table, it must always be prefixed. 
As the first two figures remain the same for a great many 
numbers in the table, they are only inserted for every ten min- 
utes, and the vacant places must be supplied from the two 
leading figures next preceding. Thus, 

on page 120, the sine of 25° 11' is 0.425516 ; 
on page 126, " " 51° 34' is 0.783332, &c. 
The tangents are found in a similar manner. Thus 
the tangent of 31° 44' is 0.618417; 
" 65° 27' is 2.18923. 

The same number in the table is both the sine of an arc and 
the cosine of its complement. The degrees for the cosines 
must be sought at the bottom of the page, and the minutes on 
the right. Thus, 

on page 130, the cosine of 16° 42' is 0.957822 ; 
on page 118, " " 73° 17' is 0.287639, &c. 

The cotangents are found in the same manner. Thus 
the cotangent of 19° 16' is 2.86089 ; 
" " 54° 53' is 0.703246. 



28 Tkigonometey. 

It is not necessary to extend the tables beyond a quadrant, 
because the sine of an angle is equal to that of its supplement 
(Art. 34). Thus 

the sine of 143° 24^ is 0.596225 ; 

" cosine of 151° 23^ is 0.877844; 
'' tangent of 132° 36: is 1.08749 ; 
" cotangent of 116° r is 0.490256, &c. 
39. If a sine is required for an arc consisting of degrees, 
minutes, and seconds, we must make an allowance for the sec- 
onds in the same manner as was directed in the case of loga- 
rithms. Art. 7; for, within certain limits, the differences of the 
sines are proportional to the differences of the corresponding 
arcs. Thus 

the sine of 34° 25^ is .565207 ; 
" " 34° 26' is .565447. 
The difference of the sines corresponding to one minute of 
arc, or 60 seconds, is .000240. The proportional part for V^ is 
found by dividing the tabular difference by 60, and the quo- 
tient, .000004, is placed at the bottom of page 122, in the col- 
umn headed 34°. The correction for any number of seconds 
will be found by multiplying the proportional part for V^ by 
the number of seconds. 

Kequired the natural sine of 34° 25' 37''. 
The proportional part for 1", being multiplied by 37, be- 
comes 148, which is the correction for 37". Adding this to 
the sine of 34° 25', we find 

the sine of 34° 25' 37" is .565355. 
Since the proportional part for 1" is given to hundredths 
of a unit in the sixth place of figures, after we have multiplied 
by the given number of seconds, we must reject the last two 
figures of the product. 

In the same manner we find 

the cosine of 56° 34' 28" is .550853. 
It will be observed, that since the cosines decrease while 
the arcs increase, the correction for the 28" is to be subtracted 
from the cosine of 56° 34'. 
In the same manner we find 

the natural sine of 27° 17' 12" is 0.458443 ; 

" " cosine of 45° 23' 23" is 0.702281 ; 



Plane Teigonometky. 29 

the natural tangent of 63° 32' 31^' is 2.0094:5; 
" " cotangent of 81° 48' 56'' is 0.14:3825. 

40. To find the niimbe7'' of degrees^ mimdes^ and seconds 
belonging to a given sine or tangent. 

If the given sine or tangent is found exactly in the table, 
the corresponding degrees will be found at the top of the page, 
and the minutes on the left hand. But when the given num- 
ber is not found exactly in the table, look for the sine or tan- 
gent which is next less than the proposed one, and take out 
the corresponding degrees and minutes. Find, also, the dif- 
ference between this tabular number and the number pro- 
posed, and divide it by the proportional part for V found at 
the bottom of the page ; the quotient will be the required 
number of seconds. 

Required the arc whose sine is .750000. 

The next less sine in the table is .749919, the arc corre- 
sponding to which is 48° 35^ The difference between this 
sine and that proposed is 81, which, divided by 3.21, gives 25. 
Hence the required arc is 48° 35' 25''. 

In the same manner we find 

the arc whose tangent is 2.00000 is 63° 26' 6". 

If a cosine or cotangent is required, we must look for the 
number in the table which is next greater than the one pro- 
posed, and then proceed as for a sine or tangent. Thus 
the arc whose cosine is .40000 is 66° 25' 18"; 
^' " " cotangent is 1.99468 is 26° 37' 34". 

41. On pages 134-5 will be found a table of natural se- 
cants for every ten minutes of the quadrant, carried to seven 
places of figures. The degrees are arranged in order in the 
first vertical column on the left, and the minutes at the top of 
the page. Thus 

the secant of 21° 20' is 1.073561 ; 
81° 50' is 7.039622. 
If a secant is required for a number of minutes not given in 
the table, the correction for the odd minutes may be found by 
means of the last vertical column on the right, which shows 
the proportional part for one minute. 

Let it be required to find the secant of 30° 33'. 
The secant of 30° 30' is 1.160592. 



30 Teigonometry. 

The correction for V is 198.9, whicli, multiplied bj 3, be- 
comes 597. Adding this to the number before found, we ob- 
tain 1.161189. 

For a cosecant, the degrees must be sought in the right- 
hand vertical column, and the minutes at the bottom of the 
page. Thus 

the cosecant of 47° 40' is 1.352742 ; 
^' " 38° 33' is 1.604626. 

42. When the natural sines, tangents, &c., are used in pro- 
portions, it is necessary to perform the tedious operations of 
multiplication and division. It is, therefore, generally prefer- 
able to employ the logarithms of the sines ; and, for conven- 
ience, these numbers are arranged in a separate table, called 
logarithmiG sines, &c. Thus 

the natural sine of 14° 30' is 0.250380. 

Its logarithm, found from page 6, is 1.398600. 
The characteristic of the logarithm is negative, as must be 
the case with all the sines, since they are less than unity. To 
avoid the introduction of negative numbers in the table, we 
increase the characteristic by 10, making 9.398600, and this 
is the number found on page 38 for the logarithmic sine of 
14° 30'. 

43. The accompanying table contains the logarithmic sines 
and tangents for every ten seconds of the quadrant. The de- 
grees and seconds are placed at the top of the page, and the 
minutes in the left vertical column. After the first two de- 
grees, the three leading figures in the table of sines are only 
given in the column headed 0", and are to be prefixed to the 
numbers in the other columns, as in the table of logarithms of 
numbers. Also, where the leading figures change, this change 
is indicated b}^ dots, as in the former table. The correction 
for any number of seconds less than 10 is given at the bottom 
of the page. 

44. To find the logarithmic sine or tangent of a given arc 
or angle. 

Look for the degrees at the top of the page, the minutes on 
the left hand, and the next less number of seconds at the top ; 
then,, under the seconds, and opposite to the minutes, will be 
found four figures, to which the three leading figures are to be 



Plane TpwIGOXometey. 31 

prefixed from the column headed 0'^ ; to this add the propor- 
tional part for the odd seconds at the bottom of the page. 
Required tlie logarithmic sine of 24:° 27' 31'^ 

The logarithmic sine of 2i° 27' 30'' is 9.61T033 

Proportional part for 4" is 18 

'' Logarithmic sine of 21° 27' 31" is 9.617051 
Required the logarithmic tangent of 73° 35' 43". 
The logarithmic tangent of 73° 35' 40" is 10.531031 
Proportional part for 3'' is 23 

Logarithmic tangent of 73° 35' 43" is 10.531054 
When a cosine is required, the degrees and seconds must be 
sought at the bottom of the page, and the minutes on the right, 
and the correction for the odd seconds must be subtracted from 
the number in the table. 

Required the logarithmic cosine of 59° 33' 47". 

The logarithmic cosine of 59° 33' 40" is 9.704682 
Proportional part for 7" is 25 

Logarithmic cosine of 59° 33' 47" is 9.701657 
So, also, the logarithmic cotangent of 37° 27' 11" is found 
to be 10.115744. 

It will be observed that for the cosines and cotangents, the 
seconds are numbered from 10" to 60", so that if it is re- 
quired to find the cosine of 25° 25' 0" we must look for 25° 
24' 60" ; and so, also, for the cotangents. 

45. The proportional parts given at the bottom of each 
page correspond to the degrees at the top of the page, in- 
creased by 30', and are not strictly applicable to any other 
num.ber of minutes; nevertheless, the differences of the sines 
change so slowly, except near the commencement of the quad- 
rant, that the error resulting from using these numbers for 
every part of the page will seldom exceed a unit in the sixth 
decimal place. Por the first two degrees, the differences 
change so rapidly that the proportional part for 1" is given 
for each minute in the right-hand column of the page. The 
correction for any number of seconds less than ten will be 
found by multiplying the proportional part for 1" by the 
given number of seconds. 

Required the logarithmic sine of 1° 17' 33". 



32 Tkigonometry. 

The logarithmic sine of 1° IT 30'' is 8.352991. 
The correction for 3'' is found by multiplying 93.4 by 3, 
which gives 280. Adding this to the above tabular number, 
we obtain for 

the sine of 1° ir 33'', 8.353271. 

A similar method may be employed for several of the first 
degrees of the quadrant, if the proportional parts at the bottom 
of the page are not thought sufficiently precise. This correc- 
tion may, however, be obtained pretty nearly by inspection, 
from comparing the proportional parts for two successive de- 
grees. Thus, on page 26, the correction for 1", corresponding 
to the sine of 2° 30', is 48 ; tlie correction for 1", correspond- 
ing to the sine of 3° 30', is 34. Hence the correction for 1", 
corresponding to the sine of 3° 0', must be about 41 ; and, in 
the same manner, we may proceed for any other part of the 
table. 

46. Near the close of the quadrant, the tangents vary so 
rapidly that the same arrangement of the table is adopted as 
for the commencement of the quadrant. For the last, as well 
as the first two degrees of the quadrant, the proportional part 
to 1" is given for each minute separatel3\ These proportional 
parts are computed for the minutes placed opposite to them, 
increased by 30", and are not strictly applicable to any other 
number of seconds ; nevertheless, the differences for the most 
part change so slowly, that the error resulting from using these 
numbers for every part of the same horizontal line is quite 
small. When great accuracy is required, the table on page 
114 may be employed for arcs near the limits of the quadrant. 
This table furnishes the differences between the logarithmic 
sines and the logarithms of the arcs expressed in seconds. 
Thus 

the logarithmic sine of 0° 5', from page 22, is 7.162696 
the logarithm of 300" (r=5') is 2.477121 

the difference is 4.685575 

This is the number found on page 114, under the heading 
log, sine A — log. J.", opposite to 5 minutes ; and, in a similar 
manner, tlie other numbers in the same column are obtained. 
These numbers vary quite slowly for two degrees ; and hence. 



Plane Teigonometky. 33 

to find the logarithmic sine of an arc less than two degrees, 
we have but to add the logarithm of the arc expressed in sec- 
onds, to the appropriate number found in this table. 
Eeqiiired the logarithmic sine of 0° 1' 22'^ 

Tabular number from page 114, 4.685575 
The lo<?arithm of 442'' is 2.645422 

Logarithmic sine of 0° T 22'' is 7.330997 
The loo:arithmic tano:ent of an arc less than two decrrees is 
found in a similar manner. 

Required the logarithmic tangent of 0° 27' 36". 
Tabular number from page 114, 4.6S5584 

The logarithm of 1656" is 3.219060 

Logarithmic tangent of 0° 27' 36" is 7.904644 
The column headed log. cot. A-\-log.A"., is found by adding 
the logarithmic cotangent to the logarithm of the arc expressed 
in seconds. Hence, to find the logarithmic cotangent of an 
arc less than two degrees, we niust subtract from the tabular 
number the logarithm of the arc in seconds. 

Required the logarithmic cotangent of 0° 27' 36". 
Tabular number from page 114, 15.314416 

The logarithm of 1656" is 3.219060 

Logarithmic cotangent of 0° 27' 36" is 12.095356 
The same method will, of course, furnish cosines and cotan- 
o^ents of arcs near 90°. 

o 

47. The secants and cosecants are omitted in this table, 
since thej are easily derived from the cosines and sines. We 
have found. Art. 35, sec. Ax cos. A = l, or taking the loga- 
rithms we have 

log. sec. A + log. COS. A = 0. 

But in the tables. Art. 42, the characteristic of the logarithm 
of each of the trigonometric functions has been increased by 
10, so that in the tables we have 

log. sec. A + log. COS. A = 20. That is, 

The logarithmic secant is found ly subtracting the loga- 
rithmic cosine from 20 ; and the logarithmic cosecant is found 
by subtracting the logarithmic sine from 20. 

Thus we have found the logarithmic sine of 24° 27' 34" to 
be 9.617051. 

C 



34 Teigonometry. 

Hence the logarithmic cosecant of 24° 27' 34^' is 10.382949. 

The logarithmic cosine of 54° 12' 40'' is 9.767008. 

Hence the logarithmic secant of 54° 12' 40" is 10.232992. 

48. To find the arc or angle corresponding to a given loga- 
rithmiG sine or tangent. 

If the given number is found exactly in the table, the cor- 
responding degrees and seconds will be found at the top of the 
page, and the minutes on the left. But when the given num- 
ber is not found exactly in the table, look for the sine or tan- 
gent which is next less than the proposed one, and take out 
the corresponding degrees, minutes, and seconds. Find, also, 
the difference between this tabular number and the number 
proposed, and corresponding to this difference, at the bottom 
of the page, will be found a certain number of seconds wdiich 
is to be added to the arc before found. 

Required the arc corresponding to the logarithmic sine 
9.750000. 

The next less sine in the table is 9.749987. 

The arc corresponding to which is 34° 13' 0"* 

The difference between its sine and the one proposed is 13, 
corresponding to whicli, at the bottom of the page, we find 4" 
nearly. Hence the required arc is 34° 13' 4". 

In the same manner, we find the arc corresponding to loga- 
rithmic tangent 10.250000 to be 60° 38' 57". 

When the arc falls within the first two degrees of the quad- 
rant, the odd seconds may be found by dividing the difference 
between the tabular number and the one proposed, by the pro- 
portional part for 1". We thus find the arc corresponding to 
logarithmic sine 8.400000 to be 1° 26' 22" nearly. 

We may employ the same method for the last two degrees 
of the quadrant when a tangent is given ; but near the limits 
of the quadrant it is better to employ the auxiliary table on 
page 114. The tabular number on page 114 is equal to log. 
sin. A— log. A". Hence log. sin. A — tabular number =: log. 
A" ; that is, if we subtract the corresponding tabular number 
on page 114, from the given logarithmic sine, the remainder 
will be the logarithm of the arc expressed in seconds. 

Required the arc corresponding to the logarithmic sine 
7.000000. 



Plane Teigonometky. 35 

"We see, from page 22, that the arc most be nearly 3^ ; the 
corresponding tabular number on page 114 is 4.685575. 

The difference is 2.314425, 
which is the logarithm of 206.''265. 

Hence the required arc is 3^ 26.^^265. 

Required the arc corresponding to log. sine 8.000000. 

We see, from page 22, that the arc is about 34^ The cor- 
responding tabular number from page 114 is 4.685568, which, 
subtracted from 8.000000, leaves 3.314432, which is the loga- 
rithm of 2062.^^68. Hence the required arc is 

34^ 22.^^68. 

In the same manner, we find the arc corresponding to loga- 
rithmic tangent 8.184608 to be 0° 52' 35''. 

Solution of Plane Right-angled Triangles. 

49. According to Art. 20, when three of the parts of a tri- 
angle are given, provided one of them is a side, the other parts 
may be determined. In a right-angled triangle, one of the 
six parts, viz. the right angle, is given ; and if one of the acute 
angles is given, the other is, of course, known. Hence in a 
right-angled triangle the number of independent parts to be 
considered is reduced to four, any two of which being given, 
the others may be found. 

It is convenient to have appropriate names by which to 
designate each of the parts of a right-angled triangle. One 
of the sides adjacent to the right angle being called the base, 
the other side adjacent to the right angle may be called the 
perpendicular. The names of the three sides will then be 
hypothenuse, base, and perpendicular. The base and perpen- 
dicular may be called the legs of the triangle. Of the two 
acute angles, that which is adjacent to the base may be called 
the angle at the base, and the other may be called the angle at 
the perpendicular. 

50. We may have four cases, according as there are given 

1. The liypothenuse and the angles; 

2. The hypothenuse and one leg ; 

3. One leg and the angles ; or 

4. The two legs. 

All of these cases may be solved by the equations of Art. 29. 



36 Teigonometrt. 

We have 




1. sin. A = cos. B=:-; 3. tan. A = cot. B 



7 7 

2. COS. A = sin. B=:-; 4. cot. A = tan.B = -. 

These equations express general truths which may be stated 
as follows : 

1. In any plane right-angled triangle, the sine of either of 
the acute angles is equal to the qitotient of the oj)posite leg di- 
vided hy the hypothenuse. 

2. The cosine of either of the acute angles is equal to the 
quotient of the adjacent leg divided hy the hypothenuse. 

3. The tangent of either of the acute angles is equal to the 
quotient of the ojpjposite leg divided by the adjacent leg. 

4. The cotangent of either of the acute angles is equal to the 
quotient of the adjacent leg divided ly the opposite leg. 

Case I. 
5 1 . Given the hyjpothenuse and the angles^ to find the two legs. 
To find a. By Eq. 1, 

sin. A == — ; whence a — c sin. A. 
c 

To find I. By Eq. 2, 

cos. A r= — : whence 1) — c cos. A. 
c 

Ex. 1. Given the hypothenuse 275, and the angle at the 

base 57° 23' to find the two legs. 

Computation hy natural nximhers. 

The natural sine of 57° 23' is 0.842296 

'' " cosine " 0.539016 

The perpendicular=rcsin. A = 275x0.842296=:231.631. 

The base = c cos. A = 275 x 0.539016 = 148.229. 

Computation hy logarithms. 

c = 275 log. 2.439333 

A==57° 23' log. sine 9.925465 

«rr:231.63 log. 2.364798 

We reject 10 from the characteristic of log. a, because the 

logarithms of the sines in the table are lew too great. 



Plane Teigonometst. 37 

c=276 log. 2.439333 

A = 57° 23^ log. COS. 9.731602 
5=148.23 log. 2.170935 

Ex. 2. Given the hypotlienuse 67.43, and the angle at the 
perpendicular 38° 43', to find the base and perpendicular. 
Ans. The base is 42.175, and perpendicular 52.612. 
The student should work this and the following examples 
both by natural numbers and by logarithms, until he has made 
himself perfectly familiar with both methods. He may then 
employ either method, as may appear to him most expeditious. 

Case II. 
52. Given the hypotlienuse and one leg^ to find the angles 
and the other leg. 

To find A. We have 



COS. A = — 

G 



To find a. We have 



sin. A = — ; whence a — c^m. A. 
c 

Ex. 1. Given the hypothenuse 54.32, and the base 32.11 to 
find the angles and the perpendicular. 

Computation by natural nuinbers. 

The quotient of 32.11 divided by 54.32 is 0.591127, which 
is the cosine of 53° 45' 17'', the angle at the base. Whence 
the other acute angle is 36° 14' 43". 

The perpendicular =rc sin. A = 54.32x 0.806580=43.813. 

The computation may be performed more expeditiously by 
logarithms, as in the former case. 

Ex. 2. Given the hypothenuse 332.49, and the perpendicu- 
lar 98.399, to find the angles and the base. 

Ans, The angles are 17° 12' 51" and 72° 47' 9" ; the base, 
317.6. 

Case III. 

53. Given one leg and the angles^ to find the other leg and 
hypothenuse. 

To find a. We have 

tan. A = -T-; whence a^h tan. A. 



38 Teigonometey. 

To find c. We have 

COS. A = — ; whence c = r-. 

c COS. A 

Ex. 1. Given the base 222, and the angle at the base 25° 15' 

to find the perpendicular and hypothenuse. 

By natural numbers we have 

the perpendicular =5 tan. A=:222x 0.471631 == 104.70; 

7) 222 

the hypothenuse = = ^ ,,,, = 215.45. 

•^ ^ COS. A 0.904455 

The computation should also be performed by logarithms, 
as in Case I. 

Ex. 2. Given the perpendicular 125, and the angle at the 
perpendicular 51° 19', to find the hypothenuse and base. 

Ans. Hypothenuse, 199.99; base, 156.12. 

Case IV. 
54. Given the two legs^ to find the angles and hypothenuse. 
To find A and B. We have 



tan. A = cot. B = -^. 





To find c. We have 



. a , a 

sm. A = — ; whence c — — 



c sm. A 

Ex. 1. Given the base 123, and perpendicular 765 to find 
the angles and hypothenuse. 
By natural numbers we have 

tan. A = ^^:= 6.219512, which is the tangent of 80° 51' 57", 
123 ' ^ 

the angle at the base. Hence the otlier acute angle is 9° 8' 3". 

The hypothenuse =: = == 774. 

•^^ sin. A 0.98732 

The computation may also be made by logarithms, as in 
Case I. 

Ex. 2. Given the base 53, and perpendicular 67, to find the 
angles and hypothenuse. 

Ans. The angles are 51° 39' 16" and 38° 20' 44" ; hypothe- 
nuse, 85,428. 



Plane Tkigonometky. 39 

55. Examjgles for Practice. 

1. Given the base 777, and perpendicular 3i5, to find the 
hjpothenuse and angles. 

This example, it will be seen, falls under Case lY. 

2. Giv^en the hypothenuse 32Jr, and the angle at the base 
48° 17^, to' find the base and perpendicular. 

3. Given the perpendicular 543, and the angle at the base 
72° 45^, to find the hypothenuse and base. 

4. Given the hypothenuse ^^^^ and base 432, to find the 
angles and perpendicular. 

5. Given the base 634, and the angle at the base 53° 27^, to 
find the hypothenuse and perpendicular. 

6. Given the hypothenuse 1234, and perpendicular 555, to 
find the base and angles. 

56. When two sides of a right-angled triangle are given, 
the third may be found by means of the property that the 
square of tlie hypothenuse is equal to the sum of the squares 
of the other two sides. 

Hence, representing the hypothenuse, base, and perpendicu- 
lar by the initial letters of these words, we have 

A=.VJM^; 1)=V¥^'', jp^y/l^^\ 
Ex. 1. If the base is 2720, and the perpendicular 3104, what 

is the hypothenuse? Ans.., 4127.1. 

Ex. 2. If the hypothenuse is 514, and the perpendicular 432, 

what is the base ? 

SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 
Theorem I. 

57. In any jplane triangle, the sines of the angles are jpro- 
portional to the opposite sides. 

jo Denote the angles of the triangle ABC 

by A, B, and C, and the sides opposite 
these angles respectively by a, 1), and c. 
From C draw CD perpendicular to AB, 
and denote CD by ^. Then in the right-angled triangles 
ACD, BCD, we have, Art. 26, 

^; = 5sin.A; _^=:<2sin.B; whence 5 sin. A = « sin. B; 
or, sin. A : sin. ~E:\a :h. 




40 



Trigonometet. 



When the angle B is obtuse, the per- 
pendicnhir falls without the triangle, and 
the angle CBD is the supplement of CBA ; 
but, by Art. 34, it has the same sine, so 
that the triangle CBD gives 
J) — asm. CBD — a sin. B; whence h sin. A = « sin. B; 
or, sin. A : sin. B : : « : J, 




the same as found when the angle B is acute. 



Whence 



sin. A sin. B sin. C' 
or, sin. A : sin. B : sin. C = a : h 



c. 



Theokem II. 

58. In any jplane triangle^ the sum of any two sides is to 
their difference^ as the tangent of ludf the sum of the ojpposite 
angles is to the tangent of half their difference. 

Let ABC be any triangle; then will 



CB + CA: CB-CA :: tan. 



A + B 



tan. 



A-B 



J 






2 2 

Produce AC to D, making CD equal to CB, and join DB. 
Take CE equal to CA, draw AE, and produce it to F. Then 
AD is the suin of CB and CA, and BE is their dfference. 

The sum of the two angles CAE, CEA, is equal to the sum 
of CAB, CBA, each being the supplement of ACB {Geora.^ 
Prop. 27, B. L). But, since CA is equal 
to CE, the angle CAE is equal to the an- 
gle CEA ; therefore, CAE is the half 
sum of the angles CAB, CBA. Also, if 
from the greater of the two angles CAB, 
CBA, there be taken their half sum, the 
remainder, Fx\B, will be their half differ- 
ence {Algebra^ p. 89). 

Since OD is equal to CB, the angle 
ADE is equal to the angle EBE; also, 
the angle CAE is equal to AEC, which 
is equal to the vertical angle BEF. Therefore, the two tri- 
angles DAF, BEF, are mutually equiangular; hence the two 
angles at F are equal, and AF is perpendicular to DB. With 
A as a centre and AF as radius describe an arc of a circle. 




Plane Teigonometrt. 41 

then DF will be the tangent of DAF, and BF will be the 
tangent of BAF. But, by similar triangles, w^e have 
AD : BE :: DF : BF; that is, 

CB + CA : CB-CA : : tan.^^^ : tan.^^^^. 



Theokem III. 

59. If from any angle of a triangle a jperjpendicular he 
drawn to the opposite side or hase, the whole hase will he to the 
sum of the other two sides, as the difference of those two sides 
is to the difference of the segments of the hase. 

For demonstration, see Geometry, Prop. 34, Gor., B. lY. 

60. In every plane triangle, three parts must be given to 
enable us to determine the others ; and of the given parts, one, 
at least, must be a side. For if the angles only are given, 
these might belong to an infinite number of different trian- 
gles. In solving oblique-angled triangles, four different cases 
may therefore be presented. There may be given, 

1. Two angles and a side ; 

2. Two sides and an angle opposite one of them ; 

3. Two sides and the included angle; or, 

4. The three sides. 

We shall represent the three angles of the proposed triangle 
by x\, B, G, and the sides opposite them, respectively, by «, &, c. 

Case I. 

61. Given two angles and a side^ to find the third angle 
and the other tv^o sides. 

To find the third angle, add the given angles together, and 
subtract their sum from 180°. 

The required sides may be found by Theorem I. The pro- 
portion will be. 

The sine of the angle opposite the given side is to the sine 
of the angle opposite the regidred side, as the given side is 
to the reguired side. C 

Ex. 1. In the triangle ABC, there are 
given the angle A, 57° 15', the angle B, ^ 
35° 30, and the side c, 364, to find the 
other parts. 




42 Tkigonometkt. 

The sura of the given angles, subtracted from 180°, leaves 
87° 15' for the angle C. Then to find the side a we have 
sin. C : sin. Kw c : a. 
By natural numbers we have 

0.998848 : 0.841039 : : 364 : ZOQA^ = a. 
This proportion is most easily worked by logarithms, thus : 
As the sine of the angle C, 87° W comp. 0.000500 
is to the sine of the angle A, 57° 15^ 9.924816 
so is the side C, 364, 2.561101 

to the side a, 306.49 2.48641T 

To find the side ^, we have 

sin. C : sin. B : : c : &. 
By natural numbers, 

0.998848 : 0.580703 : : 364 : 211.62 = 5. 
The work by logarithms is as follows : 

sin. C 87° 15' comp. 0.000500 

:sin.B 35° 30' 9.763954 

:\G 364 2.561101 

: h 211.62 2.325555 

Ex. 2. In the triangle ABC, there are given the angle A, 

49° 25', the angle C, 63° 48', and the side c, 275, to find the 

other parts. Ans., B = 66° 47'; a:=232.766; ?> = 281.67. 

Case II. 

62. Given two sides and an angle opposite one of them, to 
find the third side aiid the remaining angles. 

One of the required angles is found by Theorem I. The 
proportion is, 

The side opposite the given angle is to the side opposite the 
reguired angle, as the sine of the given angle is to the sine 
of the reqnired angle. 

The third angle is found by subtracting the sum of the other 
two from 180° ; and the third side is found as in Case I. 

If the side BC, opposite the given an- ^ 

gle A, is shorter than the other given side yf\ 

AC, the solution will be ambiguous ; that y/ V 

is, two different triangles, ABC, AB'C, y^. / \ .. 

may be formed, each of which will satisfy f^ b" B' 

the conditions of the problem. 



Plane Tkigonometet. 43 

The numerical result is also ambiguous, for the fourth term 
of the first proportion is a sine of an angle. But this may be 
the sine either of the acute angle AB'C, or of its supplement, 
the obtuse angle ABC (Art. 34). In practice, however, there 
will generally be some circumstance to determine whether the 
required angle is acute or obtuse. If the given angle is ob- 
tuse, there can be no ambiguity in the solution, for then the 
remaining angles must of course be acute. 

Ex. 1. In a triangle, ABC, there are given AC, 458, BC, 
307, and the angle A, 28° 45', to find the other parts. 
To find the angle B : 

BC : AC : : sin. A : sin. B. 
By natural numbers, 

307 : 458 : : 0.480989 : 0.717566, sin. B, 
the angle corresponding to which is 45° 51' 14'' or 134° 8' 46". 
This proportion is most easily worked by logarithms thus : 
BC, 307, comp. 7.512862 

:AC, 458, 2.660865 

::sin.A, 28° 45', 9.682135 

: sin. B, 45° 51' 14", or 134° 8' 46", 9.855862 
The angle ABC is 134° 8' 46", and the angle AB'C is 45° 
51' 14". Hence the angle ACB is 17° 6' 14'', and the angle 
ACB' is 105° 23' 46". 
To find the side AB : 

sin. A : sin. ACB : : CB : AB. 
By logarithms, 

sin. A, 28° 45', comp. 0.317865 

: sin. ACB, 17° 6' 14", 9.468502 

: : CB, 307, 2.487138 

: AB, 187.72, 2.273505 

To find the side AB' : 

sin. A : sin. ACB' : : CB' : AB'. 
By logarithms, 

sin. A, 28° 45', comp. 0.317865 

: sin. ACB', 105° 23' 46", 9.984128 
::CB', 307, 2.487138 

: AB', 615.36, 2.789131 

Ex. 2. In a triangle, ABC, there are given AB, 532, 



44 Teigonometky. 

BC, 358, and the angle C, 107° 40^ to find the other 
parts. A71S,, A .= 39° 52^ 6V' ; B=.32° 27' 9^^; AC = 299.6. 

In this example there is no ambiguity, because the given 
angle is obtuse. 

Case III. 

63. Given two sides and the included angle, to find the 
third side and the remaining angles. 

The sum of the required angles is found by subtracting the 
given angle from 180°. The difference of the required angles 
is then found by Theorem II. Half the difference added to 
half the sum gives the greater angle, and half the difference 
subtracted from half the sum gives the less angle. The third 
side is then found by Theorem I. 

Ex. 1. In the triangle ABC, the angle A is given 53° 8' ; 
the side c, 420, and the side 5, 535, to find the remaining parts. 

The sum of the angles B + C::rl80°-53° 8^=^126° 52^ 
Half their sura is 63° 26'. 

Then, by Theorem II., 
535 + 420 : 535-420 : : tan. 63° 26' : tan. 13° 32' 25", 
which is half the difference of the two required angles. 

Hence the angle B is 76° 5S' 25", and the angle C, 49° 
53' 35". 

To find the side a : 

sin. C : sin. A : : c : <3^=439.32. 

Ex. 2. Given the side c, 176, a, 133, and the included angle 
B, 73°, to find the remaining parts. 

Ans., 5 = 187.022, the angle C, 64° 9' 3", and A, 42° 50' 57". 

Case IY. 

64. Given the three sides, to find the angles. 

Let fall a perpendicular upon the longest side from the op- 
posite angle, dividing the given triangle into two right-angled 
triangles. The two segments of the base may be found by 
Theorem III. There will then be given the hypothenuse and 
one side of a right-angled triangle, to find the angles. 

Ex. 1. In the triangle ABC, the side a is 261, the side h, 
345, and c, 395. What are the angles ? 

Let fall the perpendicular CD upon AB. 



Plane Tkigonometry. 45 

Then, by Theorem III., 

AB : AC + CB : : AC-CB : AD-DB; 

or 395 : 606 : : 8i : 128.87. 

Half the difference of the se2:ments added to half their sum 
gives the greater segment, and subtracted gives the less seg- 
ment. P 

Therefore, AD is 261.935, and BD, 
133.065. V 

Then, in each of the right-angled tri- y 
angles, ACD, BCD, we have given the A d B 

hypotheniise and base, to find the angles by Case II. of right- 
angled triangles. Hence 

AD 

COS. A = ; from which A = 40° 36^ 13^^ 

AC 

cos. B=— ; " B=:59°20'52^^ 

BC' 

Heiice C = 80° 2' 55'' 

Ex. 2. If the three sides of a triangle are 150, 140, and 130, 

what are the angles ? 

Ans., 67° 22' 49'', 59° 29' 23", and 53° T 48". 

65. Examjples for Practice. 

1. Given two sides of a triangle, 478 and 567, and the in- 
cluded angle, 47° 30', to find the remaining parts. 

2. Given the angle A, 56° 34', the opposite side, a, 735, and 
the side &, 576, to find the remaining parts. 

3. Given the angle A, 65° 40', the angle B, 74° 20', and the 
side a, 275, to find the remaining parts. 

4. Given the three sides, 742, 657, and 379, to find the an- 
gles. 

5. Given the angle A, 116° 32', the opposite side, «, 492, 
and the side c, 295, to find the remaining parts. 

6. Given the angle C, 56° 18', the opposite side, c, 184, and 
the side ^, 219, to find the remaining parts. 

This problem admits of two answers. 

INSTRUMENTS USED IN DEAWTNG. 

66. The following are some of the most important instru- 
ments used in drawing. 



46 



Tkigonometky. 




I. The dividers consist of two legs, revolving upon a pivot 
at one extremity. The joints should be composed of two dif- 
ferent metals, of unequal hardness: one part, for example, of 
steel, and the other of 
brass or silver, in order 
that thejmaj move upon 
each other with greater 
freedom. The points should be of tempered steel, and when 
the dividers are closed, they should meet with great exactness. 
The dividers are often furnished with various appendages, 
which are exceedingly convenient in drawing. Sometimes one 
of the legs is furnished with an adjusting screw, by which a 
slow motion may be given to one of the points, in wliich case 
they are called hair compasses. It is also useful to have a 
movable leg, which may be removed at pleasure, and other 
parts fitted to its place; as, for example, a long beam for 
drawing large circles, a pencil point for drawing circles with 
a pencil, an ink point for drawing black circles, &c. 

67. II. The parallel rule consists of two fiat rules, made 
of wood or ivory, and connected together by two cross-bars of 




equal length, and parallel to each other. This instrument is 
useful for drawing a line parallel to a given line, through a 
given point. For this purpose, place the edge of one of the 
flat rules against the given line, and move the other rule until 
its edge coincides with the given point. A line drawn along 
its edge "will be parallel to the given line. 

68. III. Th.Q protractor is used to lay down or to measure 
angles. It consists of a semi- 
circle, usually of brass, and 
is divided into degrees, and 
sometimes smaller portions, 
the centre of tlie circle be- 
ing indicated by a small 
notch. 

To lay down an angle with the protractor, draw a base line, 




Plane Teigonometky. 



47 



and apply to it tlie edge of the protractor, so that its centre 
shall fall at the angular point. Count the degrees contained 
in the proposed angle on the limb of the circle, and mark the 
extremity of the arc with a fine dot. Remove the instrument, 
and through the dot draw a line from the angular point ; it 
will give the angle required. In a similar manner, the in- 
clination of any two lines may be measured with the pro- 
tractor. 

69. lY. The plane scale is a ruler, frequently two feet in 
length, containing a line of equal jparts^ chords^ sines, tan- 
gents, &c. For a scale of equal parts, a line is divided into 
inches and tenths of an inch, or half inches and twentieths. 
When smaller fractions are required, they are obtained by 
means of the diagonal scale, which is constructed in the fol- 
Describe a square inch, ABCD, and divide 

2 1 A .2 .4 .6 .8 B 



lowing manner. 




DE C 

each of its sides into ten equal parts. Draw diagonal lines 
from the first point of division on the upper line, to the second 
on the lower; from the second on the upper line, to the third 
on the lower, and so on. Draw, also, other lines parallel to 
AB, through the points of division of BC. Then, in the tri- 
angle ADE, the base, DE, is one tentli of an inch ; and, since 
the line AD is divided into ten equal parts, and through the 
points of division, lines are drawn parallel to the base, forming 
nine smaller triangles, the base of the least triangle is one 
tenth of DE, that is, .01 of an inch ; the base of the second is 
.02 of an inch ; the third, .03, and so on. Thus the diagonal 
scale furnishes us liundredtks of an inch. To take off from 
the scale, a line of given length, as, for example, 4.45 inches, 
place one foot of the dividers at F, on the sixth horizontal 
line, and extend the other foot to G, the fifth diagonal line. 

A half inch or less is frequently subdivided in the same 
manner. 

70. A line of chords, commonly marked ciio., is found on 
most plane scales, and is useful in setting off angles. To form 



48 



Teigonometey. 



this line, describe a circle with any convenient radius, and di- 
vide the circumference into degrees. Let the length of the 
chords for every degree of the quadrant be determined and 
laid off on a scale : this is called a line of chords. 

Since the chord of 60° is equal to radius, in order to lay 



Chords 


10 


20 


So 40 50 6o 


70 80 


90 




1 


Sines 


'? 


10 


30 40 50 6n 70 slo 


30 40 




So Secants 


6o 


TOTig. 


^ 


_4o_ 


So 40 


50 




60 





down an angle, we take from the scale the «hord of 60°, and 
with this radius describe an arc of a circle. Then take from 
the scale the chord of the given angle, and set it off upon the 
former arc. Through these two points of division draw lines 
to the centre of the circle, and they w^ill contain tlie required 
angle. 

The line of sines, commonly marked sm., exhibits the 
lengths of the sines to every degree of the quadrant, to the 
same radius as the line of chords. The line of tangents and 
the line of secants are constructed in tlie same manner. Since 
the sine of 90° is equal to radius, and the secant of 0° is the 
same, the graduation on the line of secants begins wdiere the 
line of sines ends. 

On the back side of the plane scale, are often found lines 
representing the logarithms of numbers, sines, tangents, &c. 
This is called Gunter's Scale. 

71. Y. The /Sector is a very convenient instrument in 
drawing. It consists of 
two equal arms, mova- 
ble about a pivot as a 
centre, having several 
scales drawn on the 
faces, some single, oth- 
ers double. The single scales are like those upon a common 
Gunter's scale. The double scales are those which proceed 
from the centre, each being laid twice on the same face of the 
instrument, viz., once on each leg. The double scales are a 
scale of lines, marked Lin. or L. ; the scale of chords, sines, 
&c. On each arm of the sector there is a diagonal line, which 




Plane Teigonometet. 49 

diverges from the central point like tlie radius of a circle, and 
these diagonal lines are divided into equal parts. 

The advantage of the sector is to enable us to draw a line 
upon paper to any scale ; as, for example, a scale of 6 feet to 
the inch. For this purpose, take an inch with the dividers 
from the scale of inches ; then, placing one foot of the dividers 
at 6 on one arm of the sector, open the sector until the other 
foot reaches to the same number on the other arm. Now, re- 
garding the lines on the __— ^r-xoi 

sector as the sides of a r- — ^ZXr^^S^^^^^^^^^^^^^^^^^^^''^ 

triangle, of which the /'>^ 

line measured from 6 on v^.^^ _ 

one arm to 6 on the oth- '^ ^~~------~^^^5^^?^^^¥=?=:p^^ 

er arm is the base, it is =^---~J — lq] 

plain that if any other lines be measured across the angle of 
the sector, the bases of the triangles thus formed will be pro- 
portional to their sides. Therefore, a line of 7 feet will be rep- 
resented by the distance from 7 to 7, and so on for other lines. 

The sector also contains a line of chords, arranged like the 
line of equal parts already mentioned. Two lines of chords 
are drawn, one on each arm of the sector, diverging from the 
central point. This double line of chords is more convenient 
than the single one upon the plane scale, because it furnishes 
chords to any radius. If it be required to lay down any angle, 
as, for example, an angle of 25°, describe a circle with any 
convenient radius. Open the sector so that the distance from 
60 to 60, on the line of chords, shall be equal to this radius. 
Then, preserving the same opening of the sector, place one foot 
of the dividers upon the division 25 on one scale, and extend 
the other foot to the same number upon the other scale : this 
distance will be the chord of 25 degrees, which must be set 
off upon the circle first described. 

The lines of sines, tangents, &c., are arranged in the same 
manner. 

72. By means of the instruments now enumerated, all 
the cases in Plane Trigonometry may be solved mechani- 
cally. The sides and angles which are gwe7i are laid down 
according to the preceding directions, and the reqiiired 
parts are then measured from tlie same scale. The stu- 

D 




50 Teigonometey. 

dent will do well to exercise himself upon the following 
problems : 

I. Given the angles and one side of a triangle^ to find, by 
construction, the other two sides. 

Draw an indefinite straight line, and from the scale of equal 
parts lay oS. a portion, AB, equal to the given side. From 
each extremity, lay off an angle equal to one of the adjacent an- 
gles, by means of a protractor or a scale of chords. Extend 
the two lines till they intersect, and measure tlieir lengths upon 
the same scale of equal parts which was used in laying off the 
base. 

Ex. 1. Given the angle A, 45° 30^ the 
angle B, 35° 20', and the side AB, 432 
rods, to construct the triangle, and find 
the lengths of the sides AC and BC. 

The triangle ABC may be constructed of any dimensions 
whatever; all which is essential is that its angles be made 
equal to the given angles. We may construct the triangle 
upon a scale of 100 rods to an inch, in which case the side AB 
will be represented by 4.32 inches ; or w^e may construct it 
upon a scale of 200 rods to an inch ; that is, 100 rods to a half 
inch, which is very conveniently done from a scale on which 
a half inch is divided like that described in Art. 69 ; or we 
may use any other scale at pleasure. It should, however, be 
remembered, that the required sides must be measured upon 
the same scale as the given sides. 

Ex. 2. Given the angle A, 48°, the angle C, 113°, and the 
side AC, 795, to construct the triangle. 

II. Given two sides and an angle opjposite one of tliem^ to 
find the other parts. 

Draw the side which is adjacent to tlie given angle. From 
one end of it lay off the given angle, and extend a line indefi- 
nitely for the required side. From the other extremity of the 
first side, wdth the remaining given side for radius, describe 
an arc cutting the indefinite line. The point of intersection 
will determine the third angle of the triangle. 

Ex. 1. Given tlie angle A, 74° 45', the side AC, 432, and 
the side BC, 475, to construct the triangle, and find the other 
parts. 



Plane Tkigonometrt. 



51 



Ex. 2. Given the angle A, 105°, the side BC, 498, and the 
side AC, 375, to construct the triangle. 

III. Given two sides ami the included angle ^ to find the 
other parts. 

Draw one of the given sides. From one end of it lay off 
the given angle, and draw the other given side, making the 
required angle with the first side. Then connect the extremi- 
ties of the two sides, and there will be formed the trian2:le 
required. 

Ex. 1. Given the angle A, 37° 25', the side AC, 675, and 
the side AB, 417, to construct the triangle, and find the other 
parts. 

Ex. 2. Given the angle A, 75°, the side AC, 543, and the 
side AB, 721, to construct the triangle. 

ly. Given the three sides, to find the angles. 

Draw one of the sides as a base ; and from one extremity 
of the base, with a radius equal to the second side, describe 
an arc of a circle. From the other end of the base, with a 
radius equal to the third side, describe a second arc intersect- 
ing the former; tlie point of intersection will be the third 
angle of the triangle. 

Ex. 1. Given AB, 678, AC, 598, and BC, 435, to find the 
angles. 

Ex. 2. Given the three sides 476, 287, and 354, to find the 
angles. 



Values of the Sines, Cosines, c&c, of certain Angles. 
73. We propose now to examine the changes which the 
sines, cosines, &c., undergo in the dif- 
ferent quadrants of a circle. Draw 
two diameters, AB, DE, perpendicu- 
lar to each other, and suppose one of 
them to occupy a horizontal position, 
the other a vertical. The angle ACD 
is called i\\Q first quadrant, the angle 
DCB the second quadrant, the angle 
BCE the third quadrant, and the an- 
gle ECA the fourth quadrant ; that is, the first quadrant is 
above the horizontal diameter, and on the right of the vertical 




52 Tkigonometky. 

diameter ; the second quadrant is above the horizontal diame- 
ter, and on the left of the vertical, and so on. 

Suppose one extremity of the arc remains fixed in A, while 
the other extremity, marked F, runs round the entire circum- 
ference in the direction ADBE. 

When the point F is at A, or when the arc AF is zero, the 
sine is zero. As the point F advances toward D, the sine in- 
creases ; and when the arc AF becomes 45°, the triangle CFG 
being isosceles, we have 

CG' + GF' = 2 GF^^CF^; whence GF^CF-Zf 

Therefore sin. ACF=sin. 45° = Vf 

The sine of 30° (Art. 33) is equal to one half the chord of 
60°, which is equal to the radius or unity. 

Whence sin. 30°=i=cos. 60°. 

Also, since sin. A^Vl — cos. "A, the sine of 60° which is 
equal to the cosine of 30°= Vl— i^Vf =4-'/3- 

The arc AF continuing to increase, the sine also increases 
till F arrives at D, at which point the sine is equal to the ra- 
dius or unity ; that is, the sine of 90° = 1. 

As the point F advances from D toward B, the sines dimin- 
ish, and become zero at B ; that is, the sine of 180° = 0. 

In the third quadrant, tlie sine increases again, becomes 
equal to unity at E, and is reduced to zero at A. 

74. When the point F is at A, the cosine is equal to ra-' 
dius or unity. As the point F advances toward D, the cosine 
decreases, and the cosine of 45° = sine 45°= ^^ or i'/2. The 
arc continuing to increase, the cosine diminishes till F arrives 
at D, at which point the cosine becomes equal to zero. The 
cosine in the second quadrant increases, and becomes equal to 
unity at B ; in the third quadrant it decreases, and becomes 
zero at E; in the fourth quadrant it increases again, and be- 
comes equal to unity at A. 

75. The tangent begins with zero at A, increases with the 
arc, and at 45° becomes equal to radius or unity. As the point 
F approaches D, the tangent increases very rapidly ; and when 
the difference between the arc and 90° is less than any assign- 
able quantity, the tangent is greater than any assignable quan- 
tity. Hence the tangent of 90° is said to be infinite. 



Plane Teigonometey. 



53 




In tlie second quadrant, the tangent is at first infinitely great, 
and rapidly diminishes till at B it is reduced to zero. In the 
third quadrant it increases again, becomes infinite at E, and is 
reduced to zero at A. 

The cotangent is equal to zero at D and E, and is infinite at 
A and B. 

76. The secant begins with radius or unity at A, increases 
through the first quadrant, and becomes infinite at D ; dimin- 
ishes in the second quadrant, till at B 
it is equal to unity ; increases again 
in the third quadrant, and becomes 
infinite at E ; decreases in the fourth 
quadrant, and becomes equal to unity 
at A. 

The cosecant is equal to unity at 
D and E, and is infinite at A and B. 

77. Let us now consider the al- 
gebraic signs by which these lines are to be distinguished. In 
the first and second quadrants, the sines fall above the diame- 
ter AB, while in the third and fourth quadrants they fall te- 
low. This opposition of directions ought to be distinguished 
by the algebraic signs; and if one of these directions is re- 
garded as positive, the other ought to be considered as nega- 
tive. It is generally agreed to consider those sines which fall 
above the horizontal diameter as positive ; consequently, those 
which fall below must be regarded as negative. That is, the 
sines are positive in the first and second quadrants, and nega- 
tive in the third and fourth. 

In the first quadrant the cosine falls on the right of DE, 
but in the second quadrant it falls on the left. These two lines 
should obviously have opposite signs, and it is generally agreed 
to consider those which fall to the right of the vertical diame- 
ter as positive ; consequently, those which fall to the left must 
be considered negative. That is, the cosines are positive in 
the first and fourth cjuadrants, and negative in the second and 
third. 

78. The siofns of the tangents are derived from those of 



the sines and cosines. For tan. A 



sin. A 
COS. A 



(Art. 35). Hence, 



54 Tkigonometey. 

when the sine and cosine have like algebraic signs, the tan- 
gent will be positive ; when unlike, negative. That is, the 
tangent is positive in the first and third quadrants, and neg- 
ative in the second and fourth. 

Also, cotangent A = *- (Art. 35) ; hence the tangent 

and cotangent have always the same sign. 

We have seen that sec. A = -; hence the secant must 

COS. A' 

have the same sign as the cosine. 

Also, cosec. A =-: r-; hence the cosecant must have tlie 

sin. A' 

same sign as the sine. 

79. The preceding results are exhibited in the following 

tables, w^hich should be made perfectly familiar: 

First quad. Second quad. Third quad. Fourth quad. 

Sine and cosecant, H- 4- — — 

Cosine and secant, + — — + 

Tangent and cotangent, -f — -|- — 





0° 


90° 


180° 


270° 


360° 


Sine, 





+ 1 





-1 





Cosine, 


+ 1 





-1 





+ 1 


Tangent, 





00 





00 





Cotangent, 


00 





00 





GO 


Secant, 


+ 1 


00 


-1 


00 


+ 1 


Cosecant, 


GO 


+1 


CO 


-1 


oo 



so. In Astronomy, Ave frequently have occasion to consider 
angles or arcs greater than 360°. But if an entire circumfer- 
ence, or any number of circumferences, be added to any arc, 
it will terminate in the same point as before. Hence, if C 
represent an entire circumference, or 360°, and A any arc 
whatever, we shall have 

sin.A=:sin.(C-fA)rrsin.(2C + A) = sin.(3C + A) = ,&c. 

The same is true of the cosine, tangent, &c. 

We generally consider those arcs as positive which are esti- 
mated from A in the direction ADBE. If, then, an arc were 
estimated in the direction AEBD, it should be considered as 
negative; that is, if the arc AF be considered positive, AH 



Plane Trigonometry. 



55 



must be considered negative. But the latter belongs to the 
fourth quadrant; hence its sine is negative. Therefore, sin. 
(_A) = — sin. A. 

The cosine CG is the same for both the arcs AF and AH. 
Hence, cos. (— A) = cos. A. 

Also, tan . ( - A) == - tan . A. 



And 



cot. (— A)=— cot. A, 




General Formulae. 
81. The sine and cosine of tvjo angles heing given, to find 
the sine and cosine of their sum. 

Let the two angles be AOB and BOG. 
In OC take any point C ; draw CD per- 
pendicular to OA, and CB perpendicular 
to OB. Draw BA perpendicular to OA, 
and BE perpendicular to CD. The tri- 
angles AOB and BCE are mutually equi- 
angular, since the three sides of the one are perpendicular to 
the three sides of the other respectively. Hence the angle 
BCE is equal to the angle AOB. Let the angle AOB be de- 
noted by A, and the angle BOG by B ; then the angle AOG 
will be denoted by A-f B. 

.,,-,,, CD BA-f-GE BA CE 

sm.(A + B) = ^-:.-^^- 

_BA BO CE CB 
~ BO ^ CO ■*" CB ^ CO 



Now, 



CO "^CO 



Also, 



= sin. A COS. B + cos. A sin. B. 

,, -^, OD OA-EB OA EB 

COS. (A-f B) = -^ .- 



OC 



00 OC 



OA OB EB BG 



OB ^ OC 



BC^OC 



(1) 



(2) 



= COS. A COS. B — sin. A sin. B. 

82. The sine and cosine of two angles hei7ig given, to find 
the sine and cosine of their difference. 

Let the angle AOB be denoted by A, 
and the angle BOG by B ; then the angle 
AOG will be denoted by A-B. In OC 
take any point G, draw CD perpendicular 
to OD, and CB perpendicular to OB ; 




56 Teigonometey. 

draw BE perpendicular to CD produced, and BA perpendicu- 
lar to OD. Then the triangles BOA and BEC are mutually 
equiangular, and BCE is equal to A. 

. .A -r.x CD BA-CE BA CE 

BA OB_CE CB 

"OB^OC CB^OO 

= sin. A cos. B — cos. A sin. B. (3) 

Ai /A -DN C)D OA + AD OA , BE 

Also, cos.(A-B)=^-— ^3^^-^ + ^ 

_0A OB BE CB 
~ OB' ^ OC "^ CB ^ OC 
= COS. A cos. B + sin. A sin. B. (4) 

83. From the four formulas of Arts. 81 and 82, bj addition 
and subtraction we obtain 

sin. (A + B) + sin. (A— B) = 2 sin. A cos. B (5) 
sin. (A + B)-sin. (A-B)=:2 cos. A sin. B (6) 
cos. ( A + B) + COS. (A — B) = 2 cos. A cos. B (7) 
cos. (A-B)-cos. (A + B) = 2 sin. A sin. B (8) 
If v^^e put A + Bi^C, and A — B=:D, then 

A = i-(C + D) and Br=i(C-D), and we have 
sin. C + sin. D = 2 sin. K^ + D) cos. i(C'-D) (9) 
sin. C-sin. D = 2 cos. KC + D) sin. i(C-D) (10) 
cos. C + COS. D zr: 2 cos. i(C + D) cos. i(C - D) (11) 
cos. D~cos. C=:2 sin. i(C+D) sin. HG-D) (12) 
Each of these equations expresses a general theorem. Eq. 
(9) may be enunciated thus, the sum of the sines of any two 
angles is equal to twice the sine of half the su7n of the angles 
multiplied hy the cosine of half their difference. These for- 
inuloe enable us to transform a sum or difference into a product. 

84. Since C and D admit of all varieties of value, we may 
apply the formulse (9), (10), (11), and (12) to the angles A and 
B. If, then, we divide (9) by (10) we shall have 

sin. A + sin. B _ 2 sin. -i(A + B) cos. K^- B) _ tan. i(A + B) 

sin. A-sin. B ~ 2 cos. i-(A + ^) sin. i(A- B) ~ tan. i(A-B)' 

sin. A ^ . A ^ o- 

smce 1, = tan. A. Art. 3o. 

COS. A 

This equation may be enunciated as a theorem thus : 



Plane Tkigonometet. 57 

The sum of the sines of two angles is to their difference^ as 
the tangent of half the sum of those angles is to the tangent of 
half their difference. 

85. Dividing formula (11) by (12), and remembering that 

COS. A ^ . 1 , 

T- = cot. A ~ r-? ^^^ have 

sin. A tan. A 

cos. A+cos. B _ 2 COS. ^(A + B) cos. i(A-E) _ cot. -|(A + B) ^ 
COS. B- COS. A "" 2 sin. i(AH-B) sin. ■i(A-B) ~ tan.i(A-B)' 
that is, the sum of the cosines of two angles is to their differ- 
ence, as the cotangent of half the sum of those angles is to the 
tangent of half their difference, 

86. Divide Equations (1), (2), (3), and (4) (Arts. 81 and 82) 
by cos. A COS. B, then, bj Art. 35, we have 

i^4^±^ = tan.A + tan.B (13) 
COS. A COS. B ^ ^ 

sin,(A — B) ^ A .. -n /i ^x 

^ ^5 — tan. A— tan. B (14) 

COS. A COS. B ^ ^ 

-^^?54A±^ = l-tan.Atan.B (15) 
COS. A COS. B ^ ^ 

-^5^?4^^^=:l + tan.Atan.B (16) 
COS. A COS. B ^ ^ 

87. Divide Equation (13) by 15) ; then, by Art. 35, we have 

/* , -D\ tan.A + tan.B ,.^- 

tan. (A + B) = :^ — r— ^ (17) 

^ ^ 1— tan. A tan. B ^ ^ 

Divide Eq. (14) by (16), and we obtain 

,. -p- tan. A — tan. B .._. 

tan. (A — B) = - — r— 75 (18) 

^ ^ l + tan.Atan. B ^ ^ 

by which, when the tangents of two angles are given, we may 

compute the tangent of their sum or difference. 

In the same manner we find 

, /A -TIN cot. Acot. B — 1 ,,^, 

cot.(A + B) = — — ^j- — ^ (19) 

^ ^ cot. B + cot. A ^ ^ 

cot. A cot. B + 1 .^^^ 
cot.(A-B)^ cot.B-cot.A (2^) 

88. Divide (13) by (14), Art. 86, and we have 

sin. (A + B) _ tan. A + tan. B^ 

sin. (A— B) ~ tan. A-tan. B' 

that is, the sine of the sum of two angles is to the sine of their 



58 Teigonometky. 

difference, as tlie sum of the tangents of those angles is to the 

difference of the tangents. 

Divide (15) by (16), and we have 

COS. (A + B)_l— tan. A tan.B 
COS. (A — 13) ~ 1 + tan. A tan. B* 

FormidcBfor Mtdtiple Angles. 

89. If in formulas (5) and (7), of Art. 83, we make B = A, 
we have 

sin. 2B=:2 sin.B cos. B 
COS. 2B = 2 COS. 'B-l = cos. 'B-sin. ^B (Art. 35) 
If we make A = 2B we have 

sin. 3B-j-sin. B = 2 sin. 2B cos. B, 
or, sin. 3B = 2 sin. 2B cos. B— sin. B. 
By substituting the value of sin. 2B we obtain 

sin. 3B = 4: sin. B cos. ^B— sin. B. 
Also, COS. 3B=:2 COS. 2B cos. B — cos. B, 

or, by substituting the value of cos. 2B we obtain 
cos. 3B=:4 cos/B — 3 cos. B. 
If we make A=:3B we have 

sin.4:B=:z2 sin. 3B cos.B-sin. 2B; 
or, by substitution, 

sin. 4Br^8 sin. B cos. 'B— 4 sin. B cos. B. 
Also, cos. 4B=:2 cos. 3B cos. B — cos. 2B 

= 8 cos.^B-8 cos.'B + l, 
(fee, &c. 

90. If in formula (17), Art. 87, we make Ai=B, we have 

. ^^ 2 tan. B 
tan. 2Bi= 



1-tan.^B* 
If we make A = 2B we have 

tan.2B + tan.B 



tan. 3B=: 



1 — tan. 2B tan.B* 

By substituting the value of tan. 2B we have 

^ ^-p 3 tan.B -tan. 'B 

tan. 313 — — - — -p.^ — . 

1-3 tan. 'B 

If we make A = 3B we have 

tan. 3B + tan.B 



tan.4B=: 



l-tan.3B tan.B 



J 

Plane Tkigonometkt. 59 

^^ substituting the value of tan. 3B we have 
^^ 4taii.B-4tan.^B 
^^"- ^^-1-6 tan/B + tan/B' 
&c., &c. 

91. We may also obtain the functions of the half angle in 
terms of the angle. 

If in Art. 89 we put JA for B we have 

COS. A = cos. ^|-A — sin. "-JA. 
Bj Art. 35, l = cos. "^iA + sin. ^JA. 

Subtracting the first equation from the second we have 
1-cos. A=r2 sin.'^A, 
and by addition we have 

1 + cos. A = 2 COS. '|-A. 

Hen ce, sin . i A — V^—^ cos. A, 

and cos. ^A — V^ + i- cos. A. 

92. Method of computing a table of si7ies, cosines, dbc. 

In computing a table of sines and cosines, we begin with 
finding the sine and cosine of one minute, and thence deduce 
the sines and cosines of larger angles. The sine of so small 
an angle as one minute is nearly equal to the corresponding 
arc. The radius being taken as unity, the semicircumference 
is known to be 3.14159. This being divided successively by 
180 and 60, gives .0002908882 for the arc of one minute, 
which may be regarded as the sine of one minute. 

The cosine of 1'= Vl-sm. ^1'=: 0.9999999577. 
The sines of very small angles are nearly proportional to 
the angles themselves. We might then obtain several other 
sines by direct proportion. This method will give the sines 
correct to ^ve decimal places, as far as two degrees. By the 
following method they may be obtained with greater accurac}' 
for the entire quadrant. 
In Art. 89 we found 

sin. 2B = 2 sin. B cos. B; 
sin. 3Br=2 sin. 2B cos. B-sin. B; 
sin. 4:B = 2 sin. 3B cos. B-sin. 2B, 
&c., &c. 

Making B equal to V we have 

sin.2' = 2 sin.l' cos. 1^ = 0.000582; 



60 Teigonometry. 

sin. 3^ = 2 sin. 2' cos. I'-sin. r = 0.000873 ; 
sin. 4'r=2 sin. 3' cos. 1'— sin. 2' = 0.001164, 
&c., &c. 

In like manner for the cosines we have 

cos. 2'z=2 cos. V COS. 1'— 1 =0.999999 ; 

COS. 3' = 2 COS. 2' COS. I'-cos. 1^ = 0.999999 ; 

COS. 4:^ = 2 cos. 3' COS. I'-cos. 2^ = 0.999999, 

&c., &c. 

The tangents, cotangents, secants, and cosecants are easily 

derived from the sines and cosines. Thus, 

COS. 1'^ 
shTT' 

1 



tan. r = -7; 

COS. V ' 


cot. 1^ = 


1 

sec. r — -7: 

COS. r' 


cosec. V = 


&c., 


&c. 



sin. 1' 



93. With the aid of the preceding formulae, we may obtain 
a method of solving Case 4th of oblique-angled triangles more 

Q conveniently than by the method of Art. 

/[\^^ 64. Let ABC be any plane triangle, and 

/ i \. from C draw CD perpendicular to AB. 

/ I ^x Then by Geom. B. lY., Prop. 12, 

AD B BC^ = ABM-AC^-2 ABxAD. 

Let «, b, G denote the sides opposite the angles A, B, C, then 

a' = h' + c'-2 ex AD. 
But by Art. 29, AD = & cos. A. 

Hence, (Z^ = &' + 6''— 2 he cos. A. 

That is, the square of any side of a flane triangle is equal 
to the sum of the squares of the other two sides, diminished l)y 
twice the rectangle of those sides, omdtij>lied by the cosine of 
their included angle. 

From this equation we find 

y^-^G'-a' 
'''•^= 2 be - 
by which an angle may be found when the three sides are 
given ; but this formula is not adapted to computation by 
logarithms. 

94. By Art. 91 we have 

2 sin. '^A = l — cos. A. 



Plane Teigonometey. 61 

Substituting for cos. A its value given above, we obtain 
2sm.=iA=.l ^^^ 

^{a + h — c) {a + c—h) 
^ Wg 

Put 5= the half sum of the three sides =i(« + & + c), 
then, a-\-l — c=a-{-l-{-c—^c—'^s—2c = 2{s—c)^ 

and a + c-5 = ^ + & + c- 25 = 2.5-25 = 2(5-5). 

Substituting these values in the preceding equation, we have 

sin. ^A- ^^ 



or, sin. iA =\/^ ^^ ^ (1) 

In the same manner we find 



sin.iB=V^^H5B (2) Bin.iC=\/BB (3) 

95. If it is preferred to find the angle by means of its co- 
sine, we may, by a similar method, deduce the formula 



=V^ (*) 



cos. ^A — ./ , 
'^ y OG 

Generally an angle can be determined more accurately from 
its tangent than from its sine or cosine. By a method similar 
to the preceding we obtain 

tan.iA=\/S5ES (5) 

Ex. 1. What are the angles of a plane triangle whose sides 
are 432, 543, and 654? 

Here 5 = 814.5; 5-5 = 382.5; 5-c=271.5. 

log. 382.5 2.582631 

log. 271.5 2.433770 

log. 5, 432 comp. 7.364516 

log. c, 543 comp. 7.265200 

2 ) 19.646117 
sin. -JA, 41° 42' 36.6^ 9.8230585 

Angle A = 83° 25^1 3^ 



G2 Trigonometry. 

In a similar manner we find the angle B=:4:l° 0^ od^\ and 
the angle = 55° 34^ 8^ 

Making the computation by formula (4) we have 

log. 814.5 2.910891 

log. 160.5 2.205475 

log. 432 comp. 7.361516 

log. 543 comp. 7.265200 









2) 19.746082 


COS. -1-A, 41° 


42^ 


36.9^^ 


9.873041 


By formula (5) we 


have 




log. 382.5 






2.582631 


log. 271.5 






2.433770 


log. 814.5 






comp. 7.089109 


log. 160.5 






comp. 7.794525 
2) 19.900035 



tan. -LA, 41° 42' 36.7'^ 9.9500175 

The results obtained by the three formulas differ in the 
tenths of a second. This difference arises from the use of 
logarithms to only six places. If w^e use logarithms to seven 
places, the results will agree within a tenth of a second. It 
will frequently happen, however, that the results derived from 
the three formulae will not agree as well as in this case. 

Ex. 2. What are the angles of a plane triangle whose sides 
are 245, 219, and 91? 

96. Examples/or Practice. 

Proh. 1. Given the three sides of a triangle, 627, 718.9, and 
1140, to find the angles. 

Ans., 29° 44' 2'', 34° 39' 26", and 115° 36' 32". 
Prob. 2. In the triangle ABC, the angle A is given 89° 45' 
43", the side AB 654, and the side AC 460, to find the remain- 
ing parts. Ans., BC=:798; the angle B=:35° 12' 1". and 
the angle C = 55°^ 2' 16". 
Prob. 3. In the triangle ABC, the angle A is given 56° 12' 
45", the side BC 2597.84, and the side AC 3084.33, to find the 
remaining parts. 

^^2.5.,B=:80°39'40", Cr:=43° 7'35", c=2136.8; 
or, B=:99° 20' 20," C = 24° 26' 55", c=1293.8. 



Plane Teigonometet. 63 

PtoI. 4. In the triangle ABC, the angle A is given M° 13' 
24'', the angle B 55° 59' 5S", and the side AC 368, to find the 
remainino^ parts. 

Ans^,Q=.'ir 46' 38", AB = 436.844, and BC = 309.595. 

Prob. 5. In a right-angled triangle, if the sum of the hy- 
pothenuse and base be 3409 feet, and the angle at the base 
53° 12' 14", what is the perpendicular ? Ans., 1707.2 feet. 

Proh. 6. \\i a right-angled triangle, if the difference of the 
hjpothenuse and base be 169.9 yards, and the angle at the 
base 42° 36' 12", what is the length of the perpendicular? 

Ans., 435.732 yards. 

Prob. 7. In a right-angled triangle, if the sum of the base 
and perpendicular be 123.7 feet, and the angle at the base 58° 
19' 32", what is the length of the hypothenuse? 

Ans., 89.889 feet. 

Proh. 8. In a right-angled triangle, if the difference of tlie 
base and perpendicular be 12 yards, and the angle at the base 
38° 1' 8", what is the length of the hypothennse? 

Ans., 69.81 yards. 

Prob. 9. A May-pole 50 feet 11 inches high, at a certain 
time will cast a shadow 98 feet 6 inches long; what, then, is 
the breadth of a river which runs within 20 feet 6 inches of 
the foot of a steeple 300 feet 8 inches high, if the steeple at 
tlie same time throws its shadow 30 feet 9 inches beyond the 
stream ? A7is., 530 feet 5 inches. 

Prob. 10. A ladder 40 feet long may be so placed that it 
shall reach a window 33 feet from the ground on one side of 
the street, and by turning it over, without moving the foot out 
of its place, it will do the same by a window 21 feet high on 
the other side. Kequired the breadth of the street. 

Ans., 56.649 feet. 

P?vb. 11. A May-pole was broken by a blast of wind, and 
its top struck the ground at the distance of 15 feet from the 
foot of the pole; what was the height of the whole May-pole, 
supposing the length of the broken piece to be 39 feet 1 

A71S., 75 feet. 

Prob. 12. How must three trees. A, B, C, be planted, so that 
the angle at A may be double the angle at B, the angle at B 



64 Tkigonometey. 

double the angle at C, and a line of 400 yards may just go 
round thenri ? 

Sol. Assume AB = 1, and compute the corresponding values 
of AC and BC. 

Ans,, AB = 79.225, AC ^ 142.758, and BC =: 178.017 yards. 

Prob. 13. The town B is half-way between the towns A 
and C, and the towns B, C, and D are equidistant from each 
other. What is the ratio of the distance AB to AD ? 

Ans.^ As unity to -y/Z. 

Prob. 14. There are two columns left standing upright in 
the ruins of Persepolis; the one is QQ feet above the plain, 
and the other 48. In a straight line between them stands an 
ancient statue, the head of which is 100 feet from the summit 
of the higher, and 84 feet from the top of the lower column, 
the base of which measures just 74 feet to the centre of the 
figure's* base. Kequired the distance between the tops of the 
two columns. Ans., 156.68 feet. 

Prol). 15. Prove that tan. (45°- A) = 1'^^' ^ - 

^ ^ 1-f-tan. A 

Prob. 16. One angle of a triangle is 45°, and the perpendic- 
ular from this angle upon the opposite base divides the base 
into two parts, which are in the ratio of 2 to 3. What are 
the parts into which the vertical angle is divided by this per- 
pendicular? 

Sol. Let cc = the larger angle; then 

tan. (45°-A)=f tan. A=r ^"",^^"- f , 
^ ^ ^ 1 + tan. A' 

which can be solved as an equation of the second degree. 

Ans., 18° 26' &', and 26° 33' 54''. 

Prob. 17. Prove that sin. 3A=:3 sin. A— 4 sin. 'A. 

Prob. 18. One side of a triangle is 25, another is 22, and 
the angle contained by these two sides is one half of the angle 
opposite the side 25. What is the value of the included angle ? 

sin. ?>x 3 sin. a? — 4 sin. ^x 3 — 4 sin. ""x 

boo. — — TT- = .bo = — -^ — -. = — -^ — 

sm. 2a? 2 sm. x cos. x ^ cos. x 

which can be solved as an equation of the second 



2-v/l-sin. 

degree. Ans., 39° 58' 51^ 



Plane Teigonometet. 65 

Prob. 19. One side of a triangle is 25, another is 22, and 
the angle contained by these two sides is one half of the angle 
opposite the side 22. AYhat is the value of the included angle ? 
Sol. Like the preceding. Ans., 30° 46' 38''. 

Prob. 20. Two sides of a triangle are in the ratio of 11 to 
9, and the opposite angles have the ratio of 3 to 1. What are 
those angles ? 
Sol. 3 sin. 0?— 4 sin. ^x : sin. a? : : 11 : 9. 

Ans. The sine of the smaller of the two angles is f , and 

of the greater ff ; the angles are 41° 48' 37", and 

125° 25' 51". 

Prob. 21. One side of a triangle is 15, and the difference of 

the two other sides is 6 ; also, the angle included between the 

first side and the greater of the two others is 60°. What is 

the length of the side opposite to this angle ? Ans.^ 57. 

Prob. 22. One side of a triangle is 15, and the difference of 
the two other sides is 6 ; also, the angle opposite to the greater 
of the two latter sides is 60°. What is the length of said side? 

Ans., 13. 
Prob. 23. One side of a triangle is 15, and the opposite an- 
gle is 60° ; also, the difference of the two other sides is 6. 
What are the lengths of those sides ? 

Ans., 11.0712, and 17.0712. 
Prob. 24. The perimeter of a triangle is 100 ; the perpen- 
dicular let fall from one of the angles upon the opposite base 
is 30, and the angle at one end of this base is 50°. What is 
the length of the base ? Ans., 30.388. 

E 



CHAPTER III. 

MENSURATION OF SURFACES. 

97. The area of a figure is the space contained within the 
line or lines by which it is bounded. This area is determined 
by finding how many times the figure contains some other 
surface, which is assumed as the unit of measure. This unit 
is commonly a square ; such as a square inch, a square foot, a 
square rod, &c. 

The superficial unit has generally tlie same name as the 
linear unit, which forms the side of the square. Thus, 
the side of a square inch is a linear inch ; 
" " of a square foot is a linear foot ; 
" " of a square yard is a linear yard, &c. 

There are some superficial units which have no correspond- 
ing linear units of the same name, as, for example, an acre. 

The following table contains the square measures in com- 
mon use: 





Table of Sq 


uare Measures. 


Sq. Inches. 


Sq. Feet. 




, 144 = 


1 sq. 


Yards. 


1296 = 


9 = 


1 Sq. Rods. 


39204 = 


272^= 


30 J = 1 Sq.Ch's. 


627264 = 


4356 = 


484 = 16= 1 ^,res. 


6272640 = 


43560 = 


4840 = 160= 10= 1 M. 



4014489600 = 27878400 =3097600 =102400 = 6400 = 640 = 1 

Problem I. 
98. To find the area of a parallelogram. 

Rule I. 

Multiply the lase hy the altitude. 

For the demonstration of this rule, see Geometry, Prop. 5, 
B.IY. 



Mensuration of Surfaces. 67 

Ex. 1. What is the area of a parallelogram wliose base is 
17.5 rods, and the altitude 13 rods ? 

Ans., 227.5 square rods. 
Ex. 2. What is the area of a square whose side is 315.7 
feet ? Ans., 99666.49 square feet. 

Ex. 3. What is the area of a rectangular board whose length 
is 15.25 feet, and breadth 15 inches? 

A71S., 19.0625 square feet. 
Ex. 4. How many square yards are there in the fom* sides 
of a room which is 18 feet long, 15 feet broad, and 9 feet high? 

Ans., 6Q square yards. 

99. If the sides and angles of a parallelogram are given, 
the perpendicular height may be found by 
Trigonometry. For DE is one side of a 
right-angled triangle, of which AD is tlie / ; / 
hypothenuse. / J / 

By Art. 50 we have ^ ^ ^ 

DE = ADxsin. A. 
Therefore the area = AB x BE z=ABx AD x sin. A. 
Hence we derive 

EULE II. 

Multiply together two adjacent sides, and the sine of the iti- 
cluded angle. 

Ex. 1. What is the area of a parallelogram having an angle 
of 58°, and the including sides 36 and 22.5 feet? 

Ans. The area = 36 x 25.5 x .81805 (natural sine of 58°) = 
778.508 square feet. 

The computation Tvill generally be most conveniently per- 
formed by logarithms. 

Ex. 2. What is the area of a rhombus, each of whose sides 
is 21 feet 3 inches, and each of the acute angles 53° 20^ ? 

Ans., 362.209 feet. 

Ex. 3. Plow many acres are contained in a parallelogram 
one of whose angles is 30°, and the including sides are 25.35 
and 10.4 chains? Ans., 13 acres, 29.12 rods. 

Problem II. 

100. To find the area of a triangle. 



68 Tkigonometkt. 

Rule I. 

Multiply the hase ly half the altitude. 

For demonstration, see Geometry, Prop. 6, B. lY. 

Ex. 1. How many square yards are contained in a triangle 
whose base is 49 feet, and altitude 25J feet ? 

Ans., 68.736. 

Ex. 2. What is the area of a triangle whose base is 45 feet, 
and altitude 27.5 feet? Ans.^ 618.75 square feet. 

101. When two sides and the included angle are given, we 
may use 

Rule II. 
Multiply half thefroduct of two sides ly the sine of the in- 
cluded angle. 

The reason of this rule is obvious, from Art. 99, since a tri- 
angle is half of a parallelogram, having the same base and al- 
titude. 

Ex. 1. What is the area of a triangle of which two sides are 
45 and 32 feet, and the included angle 46° 30' ? 

Ans. The area = 45 x 16 x .725374 (natural sine of 46° 30') = 

522.269 feet. 
Ex. 2. What is the area of a triangle of which two sides are 
127 and 96 feet, and the included angle 67° 15' ? 

Ans. 

102. When the three sides are known, we may use 

Rule HI. 

From half the sum of the three sides suttraet each side sev- 
erally ; multiply together the half sum and the three remain- 
ders^ and extract the square root of the product. 

Demonstration. 
Let a^ 5, c denote the sides of the triangle ABC ; then, by 
Geometry, Prop. 12, B. lY., we have BC = c 

AB^4-AC^-2ABxAD,or a^ = 5Hc^-2(j /K 

xAD; whence, / \ \ 

2C A D B 



Mexsueation of Stjefaces. 69 

But CD^ = AC^-AD^; 

hence, CD =h -^, = ^, ; 



or, \jiJ = 



2c 



AB X CD 



But the area= ^ =:X^±lf(f-{(f + (f-ary. 

The quantity under the radical sign being the difference of 
two squares, may be resolved into the factors 2hc + Qf + c'^ — ar) 
and 2Jc— (5' + 6''— «') ; and these, in the same manner, may be 
resolved into {h + c + a) x{h + c — a), 

and {a+b-c)x{a — h + c). 

Hence, if we put S equal to — ^ , we shall have 



the area=VS{S-a) (S—b) (S-c). 
Ex. 1. What is the area of a triangle whose sides are 125, 
173, and 216 feet? 
Here S-257, S-5==Si, 

S-f^ = 132, S-c = 4:l. 

Hence the area= 1/257 x 132 x 81 x il = 10S09 square feet. 

Ex. 2. How many acres are contained in a triangle whose 
sides are 49, 50.25, and 25.69 chains? 

Ans., 61 acres, 1 rood, 39.68 perches. 

Ex. 3. What is the area of a triangle whose sides are 234, 
289, and 345 feet? Ans. 

103. In an equilateral triangle, one of whose sides is a, the 
expression for the area becomes 

V^a x^ax ^a x ^a 

~ 4 ' 
that is, the area of an equilateral triangle is equal to one 
fourth the square of one of its sides multiplied by the square 
root of 3. 

Ex. What is the area of a triangle whose sides are each 37 
feet ? A71S., 592.79 feet. 

Peoblem hi. 

104. To find the a rea of a trajpezoid. 



70 Tkigonometey. 

EULE. 

Midtiply half the su7n of the parallel sides into their jper- 
pendicular distance. 

For demonstration, see Geometry, Prop. 7, B. lY. 

Ex. 1. What is the area of a trapezoid whose parallel sides 
are 156 and 124, and the perpendicular distance between them 
57 feet ? Ans., 7980 feet. 

Ex. 2. How many square yards in a trapezoid whose par- 
allel sides are 678 and 987 feet, and altitude 524 feet ? 

Ans. 

Peoblem IY. 

105. To find the area of an irregidar polygon. 

Ktjle. 

Draw diagonals dividing the polygon into triangles^ and 
find the sum of the areas of these triangles. 

Ex. 1. What is the area of a quadrilateral, one of whose 
diagonals is 126 feet, and the two perpendiculars let fall upon 
it from the opposite angles are 74 and 28 feet % 

Ans., 6426 feet. 

Ex. 2. In the polygon ABODE, there 
are given EC=:205, EB = 242, AF = 65, 
CG=114,and DH=.110,tofind the area. 
Ans. 

106. If the diagonals of a quadrilateral 
are given, the area may be found by the 
following 

RlJLE. 

Multiply half the 'product of the diagonals hy the sine of 
the angle at their intersection. 

Demonstration. 
The sines of the four angles at E are all 
equal to each other, since the adjacent an- 
gles AED, DEC are the supplements of 
each other (Art. 34). But, according to 
the Kule, Art. 101, the area of 





Mensijkation of Suefaces. 71 

the triangle ABE = |-AExBE xsine E; 

'' " AED^iAExDExsiiieE; 

" " BEC^iBExECxsineE; 

" " DEC^iDExECxsineE. 

Therefore, 

the area of ABCD^K^E + EC) x(BE + ED) xsine E 

=i-ACxBDxsineE. 
Ex. 1. If the diagonals of a quadrilateral are 34 and 56 
rodSj and if they intersect at an angle of 67°, what is the area? 

Ans., 876.32. 
Ex. 2. If the diagonals of a quadrilateral are 75 and 49, and 
the angle of intersection is 42°, what is the area ? 

Ans. 
Problem Y. 

107. To find the area of a regular jpolygon. 

Rule I. 
Multiply half the perimeter hy the perpendicular let fall 
from the centre on one of the sides. 

For demonstration, see Geometry, Prop. 7, B. YI. 
Ex. 1. What is the area of a regular pentagon whose side 
is 25 feet, and the perpendicular from the centre 17.205 feet? 

Ans., 1075.31 feet. 
Ex. 2. What is the area of a regular octagon whose side is 
53, and the perpendicular 63.977? Ans. 

108. When the perpendicular is not given, it may be com- 
puted from the perimeter and number of sides. If we divide 
360 degrees by the number of sides of the 
pol3\gon, the quotient will be the angle / 
ACB at the centre, subtended by one of / 
the sides. The perpendicular CD bisects \ 
the side AB, and the angle ACB. Then, \ 
in the triangle ACD, we have (Art. 50), 

CD = ADcot.ACD. 

The perpendicular from the centre is equal to half of one 
of the sides of the polygon multiplied ly the cotangent of the 
opposite angle. 

Ex. 3. Find the area of a regular hexagon whose side is 32 
inches. 



D 



72 Tkigonometey. 

The angle ACD is ^V of 360° = 30°. Then 

CD, the perpendicularrzrlGxcot. 30° = 27.7128, 
and the area = 27.7128x 16 x 6 ==: 2660.4288. 
Ex. 4. Find the area of a regular decagon whose side is 46 
feet. Ans., 16280.946. 

109. In this manner was compnted the following table of 
the areas of regular polygons, in which the side of each poly- 
gon is supposed to be a unit. 



Table of Kegijlae 


Polygons. 


Xames. 


Sides. 


Areas. 


Triangle, 


3 


0.4330127. 


Square, 


4 


1.0000000. 


Pentagon, 


5 


1.7204774. 


Hexagon, 


6 


2.5980762. 


Heptagon, 


7 


3.6339124. 


Octagon, 


8 


4.8284271. 


Nonagon, 


9 


6.1818242. 


Decagon, 


10 


7.6942088. 


Undecagon, 


11 


9.3656399. 


Dodecao^on, 


12 


11.1961524. 



By the aid of this table may be computed the area of any 
other regular polygon having not more than twelve sides. For, 
since the areas of similar polygons are as the squares of their 
homologous sides, we derive 

EULE II. 

Multiply the square of one of the sides of the polygon ly the 
area of a similar polygon whose side is unity. 

Ex. 5. What is the area of a regular nonagon whose side 
is 63? ' ^^15., 24535.66. 

Ex. 6. What is the area of a regular dodecagon whose side 
is 54 feet ? Ans., 32647.98 feet. 

Problem YI. 
110. To find the circumference of a circle from its diameter. 

Rule. 
Multip)ly the diameter Jy 3.14159. 



Meitsueation of Surfaces. 73 

For the demonstration of this rule, see Geometry, Prop. 13, 
Cor. 2, B. YI. 

When the diameter of the circle is small, and no great ac- 
curacy is required, it may be sufficient to employ the value 
of TT to only 4 or 5 decimal places. But if the diameter is 
large, and accuracy is required, it will be necessary to employ 
a corresponding number of decimal places of tt. The value 
of - to ten decimal places is 3.14159,26536, 

and its logarithm is 0.497150. 

Ex. 1. What is the circumference of a circle whose diameter 
is 125 feet ? Ans., 392.7 feet. 

Ex. 2. If the diameter of the earth is 7912 miles, what is its 
circumference? Ans., 24856.28 miles. 

Ex. 3. If the diameter of the earth's orbit is 189,761,000 
miles, what is its circumference ? Ans., 596,151,764 miles. 

To obtain this answer, the value of tt must be taken to at 
least eight decimal places. 

Peoblem YII. 

111. To find the diameter of a circle from its circum- 
ference. 

Rule I. 

Divide the circumference hj 3.14159. 

This rule is an obvious consequence from the preceding. 
To divide by a number is the same as to multiply by its re- 
ciprocal ; and, since multiplication is more easily performed 
than division, it is generally most convenient to multipl}^ by 
the reciprocal of tt, which is 0.3183099. Hence we have 

Pule II. 

Multiply the circumference hy 0.31831. 

Ex. 1. What is the diameter of a circle whose circumference 
is 875 feet ? Ans., 278.52 feet. 

Ex. 2. If the circumference of the moon is 6786 miles, what 
is its diameter? Ans.^ 2160 miles. 

Ex. 3. If the circumference of the moon's orbit is 1,492,987 
miles, what is its diameter? Ans.^ 475,233 miles. 



74 Tkigonometry. 

Problem YIII, 

112. To find the length of an arc of a circle, 

KULE I. 

As 360 is to the numter of degrees in the arc^ so is the cir- 
cumference of the circle to the length of the arc. 

This rule follows from Prop. 14, B. III., in Geometry, where 
it is proved that angles at the centre of a circle have the same 
ratio with the intercepted arcs. 

Ex. 1. What is the length of an arc of 22°, in a circle whose 
diameter is 125 feet? 

The circumference of the circle is found to be 392.7 feet. 
Then 360 : 22 : : 392.7 : 23.998 feet. 

Ex. 2. If the circumference of the earth is 24,856.28 miles, 
what is the length of one degree ? Ans.^ 69.045 miles. 

Pule II. 

113. Multiply the diameter of the circle hy the nwnber of 
degrees in the arc, and this product ly 0.0087266. 

Since the circumference of a circle whose diameter is unity 
is 3.14159, if we divide this number by 360, we shall obtain 
the length of an arc of one degree, viz., 0.0087266. If we 
multiply this decimal by the number of degrees in any arc, we 
shall obtain the lens^th of that arc in a circle whose diameter 
is unity ; and this product, multiplied by the diameter of any 
other circle, will give the length of an arc of the given num- 
ber of degrees in that circle. 

Ex. 3. What is the length of an arc of 25°, in a circle whose 
radius is 44 rods ? Ans., 19.198 rods. 

Ex. 4. What is the length of an arc of 11° 15', in a circle 
whose diameter is 1234 feet ? Ans., 121.147 feet. 

114. If the number of degrees in an arc 
is not given, it may be computed from the 
radius of the circle, and either the chord 
or height of the arc. Thus, let AB be the 
chord, and DE the height of the arc ADB, 
and C the centre of the circle. Then, in 
the right-angled triangle ACE, 




Mensuration of Suefaces. 75 



siD. ACE=:^-^; COS. ACE =3—-; 

either of which equations will give the number of degrees 
in half the arc. 

If only the chord and height of the arc are given, the diam- 
eter of the circle may be found. For, by Geometry, Prop. 23, 
Cor., B. lY., 

DE : AE : : AE : EF. 

Ex. 5. What is the length of an arc whose chord is 6 feet, 
in a circle whose radius is 9 feet? A71S., 6.117 feet. 

Peoblem IX. 

115. To find the area of a circle. 

KULE I. 

Multiply the circumference hy half the radius. 
For demonstration, see Geometry, Prop. 12, B. YI. 

ErLE II. 

MuUijoly the square of the radius hy 3.14159. 

See Geometry, Prop. 13, Cor. 3, B. YI. 

Ex. 1. What is the area of a circle whose diameter is 18 
feet ? Ans., 254.469 feet. 

Ex. 2. What is tlie area of a circle whose circumference is 
74 feet ? Ans., 435.766 feet. 

Ex. 3. What is the area of a circle whose radius is 125 
yards ? A71S., 490S7.3S yards. 

Peoblem X. 

116. To find the area of a sector of a circle. 

Pule I. 

Multiply the arc of the sector hy half its radius. 
See Geometry, Prop. 12, Cor., B. YI. 

EULE II. 
As 360 is to the numher of degrees in the arc, so is the area 
of the circle to the area of the sector. 
This follows from Geometry, Prop. 14, Cor., B. III. 



T6 Trigokometey. 

Ex. 1. What is the area of a sector whose arc is 22°, in a 
circle whose diameter is 125 feet ? 

The length of the arc is found to be 23.998. 

Hence the area of the sector is 749.937. 

Ex. 2. What is the area of a sector whose arc is 25°, in a 
circle whose radius is M rods ? A7is., 422.367 rods. 

Ex. 3. What is the area of a sector less than a semicircle, 
whose chord is 6 feet, in a circle whose radius is 9 feet ? 

Ans., 27.522. 

Peoblem XI. 

117. To find the area of a segment of a circle. 

EULE. 

Find the area of the sector which has the same arc, and also 
the area of the triangle formed hy the chord of the segment and 
the radii of the sector. 

Then take the sum of these areas, if the segment is greater 
than a semicircle, hut take their difference if it is less. 

It is obvious that the segment AEB is 
equal to the sum of the sector ACBE and 
the triangle ACB, and that the segment 
ADB is equal to the difference between 
the sector ACBD and the triangle ACB. 

Ex. 1. What is the area of a segment 
whose arc contains 280°, in a circle whose 
diameter is 50 ? 

The whole circle = 1963.495 
The sector = 1527.163 

The triangle = 307.752 

The segment =2 1834.915, Ans. 

Ex. 2. What is the area of a segment whose chord is 20 feet, 
and height 2 feet ? Ans., 26.8788 feet. 

Ex. 3. What is the area of a segment whose arc is 25°, in a 
circle whose radius is 44 rods ? Ans. 

118. The area of the zone ABPIG, included between two 
parallel chords, is equal to the difference between the segments 
GDH and ADB. 




Mensuration of Sukfaces. 77 

Ex. 4. What is the area of a zone, one side of which is 96, 
and the other side 60, and the distance between them 26 ? 

A?is., 2136.7527. 

The radius of the circle in this example will be found to 
be 50. 

Peoblem XII. 

119. To find the area of a ring included behucen the cir- 
cumferences of tico concentric circles. 

Rule. 
Tcike the difference between the areas of the two circles / or, 
Subtract the square of the less radius from the square of the 
greater^ and midtij)ly their difference by 3.14:159. 
For, according to Geometry, Prop. 13, Cor. 3, B. YL, 
the area of the greater circle is equal to it H', 
and the area of the smaller, tt r'. 

Their difference, or the area of the ring, is - {K' — r'). 
Ex. 1. The diameters of two concentric circles are 60 and 
50. What is the area of the rins^ included between their cir- 
cumferences? J.n6'., 863.938. 

Ex. 2. The diameters of two concentric circles are 320 and 
280. What is the area of the ring included between their 
circumferences? Ans.^ 18849.55. 

Peoblem XIII. 

120. To find the area of an ellijpse. 

Rule. 

Multiply the ^product of the semi-axes by 3.14159. 

For demonstration, see Geometry, Ellipse, Prop. 23. 

Ex. 1. What is the area of an ellipse whose major axis is 
70 feet, and minor axis 60 feet ? Ans., 3298.67 feet. 

Ex. 2. What is the area of an ellipse whose axes are 340 
and 310? ^ns., 82780.896. 

Peoblem XIY. 

121. To find the area of a parabola. 



78 Tkigonometey. 

Exile. 

Multiply the base hy two thirds of the height. 

For demonstration, see Geometry, Parabola, Prop. 16. 

Ex. 1. What is the area of a parabola whose base is 18 feet, 
and height 5 feet ? Ans., 60 feet. 

Ex. 2. What is the area of a parabola whose base is 525 
feet, and height 350 feet ? Ans., 122500 feet 

MENSURATION OF SOLIDS. 

122. The common measuring unit of solids is a cube, 
whose faces are squares of the same name ; as, a cubic inch, 
a cubic foot, &c. This measuring unit is not, however, of 
necessity a cube, whose faces are squares of the same name. 
Thus a bushel may have the form of a cube, but its faces can 
only be expressed by means of some unit of a different denom- 
ination. The following is 

The Table of Solid Pleasure. 
1728 cubic inches = 1 cubic foot. 
27 cubic feet = 1 cubic yard. 
44921 cubic feet = 1 cubic rod. 
231 cubic inches = 1 gallon (liquid measure). 
268.8 cubic inches = 1 gallon (dry measure). 
2150.4 cubic inches = 1 bushel. 

Problem I. 

123. To find the surface of a right ijrism. 

EULE. 

Multiply the perimeter of the base by the altitude, for the 
convex surface. To this add the areas of the two ends when 
the entire surface is required. 

See Geometry, Prop. 1, B. YIII. 

Ex. 1. What is the entire surface of a parallelepiped whose 
altitude is 20 feet, breadth 4 feet, and depth 2 feet % 

Ans.^ 256 square feet. 

Ex. 2. What is the entire surface of a pentagonal prism 
whose altitude is 25 feet 6 inches, and each side of its base 3 
feet 9 inches ? Ans., 526.513 square feet. 



Mensuration of Solids. 79 

Ex. 3. What is the entire surface of an octagonal prism 
whose altitude is 12 feet 9 inches, and each side of its base 2 
feet 5 inches? Ans., 302.898 square feet. 

Problem II. 

124. To find the solidity of a prism. 

KULE. 

Multijply the area of the lase hj the altitude. 
See Geometry, Prop. 11, B. YIII. 

Ex. 1. What is the solidity of a parallelepiped whose alti- 
tude is 30 feet, breadth 6 feet, and depth 4 feet? 

Ans., 720 cubic feet. 
Ex. 2. What is the solidity of a square prism whose altitude 
is 8 feet 10 inches, and each side of its base 2 feet 3 inches? 

Ans., 4:4:f f cubic feet. 
Ex. 3. What is the solidity of a pentagonal prism whose al- 
titude is 20 feet 6 inches, and its side 2 feet 7 inches? 

A71S., 235.376 cubic feet. 

Problem III. 

125. To find the surface of a regular pyramid. 

Rule. 

Ilidtiply the perimeter of the lase ly half the slant height 
for the convex surface. To this add the area of the lase when 
the entire surface is required. 

See Geometry, Prop. 14, B. YIII. 

Ex. 1. What is the entire surface of a triangular pyramid 
whose slant height is 25 feet, and each side of its base 5 feet? 

Ans., 198.325 square feet. 
Ex. 2. What is the entire surface of a square pyramid whose 
slant height is 30 feet, and each side of the base 4 feet ? 

Ans., 256 square feet. 
Ex. 3. What is the entire surface of a pentagonal pyramid 
whose slant heiglit is 20 feet, and each side of the base 3 feet? 

A')is., 165.484 square feet. 



80 Teigonometey. 

Pkoblem IY. 

126. To find the solidity of a pyramid. 

Rule. 

Midtiply the area of the lase hj one third of the altitude. 
See Geometry, Prop. 17, B. YIIL 

Ex. 1. What is the solidity of a triangular pyramid whose 
altitude is 25 feet, and each side of its base feet ? 

Ans., 129.904 cubic feet. 
Ex. 2. What is the solidity of a square pyramid whose slant 
height is 22 feet, and each side of its base 10 feet ? 

Ans., 714.143 cubic feet. 
Ex. 3. What is the solidity of a pentagonal pyramid whose 
altitude is 20 feet, and each side of its base 3 feet ? 

Ans., 103.228 cubic feet. 

Problem Y. 

127. To find the surface of a frustum of a regular pyr- 
amid. 

Rule. 

Multiply half the slant height hy the sum of the perimeters 
of the two hasesfor the convex surface. To this add the areas 
of the two hases when the entire surface is reguired. 

See Geometry, Prop. 14, Cor. 1, B. YIII. 

Ex. 1. What is the entire surface of a frustum of a square 
pyramid whose slant height is 15 feet, each side of the greater 
base being 4 feet 6 inches, and each side of the less base 2 feet 
10 inches ? Ans.^ 248.278 square feet. 

Ex. 2. What is the entire surface of a frustum of an octag- 
onal pyramid whose slant height is 14 feet, and the sides of 
the ends 3 feet 9 inches, and 2 feet 3 inches? 

Ans.^ 428.344 square feet. 

Peoblem YI. 

128. To find the solidity of a frustu m of a pyramid. 



Menstjeation of Solids. 



81 



EULE. 

Add together the areas of the two hases^ and a mean j^rojpor- 
tional between them^ and multiply the sum hy one third of the 
altitude. 

See Geometry, Prop. 18, B. YIIL, Scholium. 

When the pyramid is regular, it is generally most conven- 
ient to find the area of its base by Rule II., Art. 109. If we 
put a to represent one side of the lower base, and h one side 
of the upper base, and the tabular number from Art. 109 by 
T, the area of the lower base will be a'T ; that of the upper 
base will be Z'^T ; and the mean proportional will be dbH. 
Hence, if we represent the height of the frustum by A, its so- 
lidity will be 

(a^^V'^ab)^. 

Ex. 1. What is the solidity of a frustum of an hexagonal 
pyramid whose altitude is 15 feet, each side of the greater end 
being 3 feet, and that of the less end 2 feet ? 

Ans., 24:6.817 cubic feet. 
Ex. 2. What is the solidity of a frustum of an octagonal 
pyramid whose altitude is 9 feet, each side of the greater end 
being 30 inches, and that of the less end 20 inches \ 

Ans., 191.125 cubic feet. 



129. 



Definition. 
A wedge is a solid bounded by five planes, viz. 



a rec- 



tangular base, ABCD, two trape- 
zoids, ABFE, DCFE, meeting in 
an edge, and two triangular ends, 
ADE, BCF. The altitude of the 
wedge is the perpendicular drawn 
from any point in the edge to the 
plane of the base, as EH. 




130. 



Peoblem YII. 
To find the solidity of a wedge. 



Rule. 

Add the length of the edge to twice the length of the lase, 

F 



S% Tkigonometey. 

a7id multiply the sum hy one sixth of tlte jproduct of the height 
of the wedge and the 'breadth of the base. 

Demonstration. 
Put L — AB, the length of the base; 
'' Z = EF, the length of the edge ; 
" J = BC, the breadth of the base ; 
" /f =:EH, the altitude of the wedge. 
Now, if the length of the base is equal to that of the edge, 
it is evident that the wedge is half of a prism of tlie same base 
and height. If the length of the base is greater than that of 
the edge, let a plane, EGI, be drawn parallel to BCF. The 
wedge will be divided into two parts, viz., the pyramid E — 
AIGD, and the triangular prism BCF— G. 

The solidity of the former is equal to ^bh(L — I), and that of 
the latter is ^hhl. Their sum is 

ihhl + ihh{L-T) ^ ihhSl + ihh2L - lhh2l = iU{2L + 1). 
If the length of the base is less than that of the edge, the 
wedge will be equal to the difference between tlie prism and 
pyramid, and we shall have 

^hhl-ihh{l-L), 
which is equal to 

ilhl + ihh{L-l\ 
the same result as before. 

Ex. 1. What is the solidity of a wedge whose base is 30 
inches long and 5 inches broad, its altitude 12 inches, and the 
length of the edge 2 feet? Ans., 840 cubic inches. 

Ex. 2. What is the solidity of a wedge whose base is 40 
inches long and 7 inches broad, its altitude 18 inches, and the 
length of the edge 30 inches? Ans., 2310 cubic inches. 

Definition. 
131. A rectangular jprismoid is a solid bounded by six 
planes, of which the two bases are rectangles having their 
corresponding sides parallel, and the four upright sides of the 
solids are trapezoids. 

Pkoblem YIII. 

To find the solidity of a rectangular jyrismoid. 



Mensuration of Solids. 83 

EULE. 

Add together the areas of the two hases, and four times the 
area of a parallel section equally distant from the lases^ and 
midtiply the sum hy one sixth of the altitude. 

Demonstration. 
Put L and B=lengtli and breadth of one base ; 
Put I and &= length and breadth of tlie other base ; 
" M " m=length and breadth of middle sec; 
" h =: the altitude of the prismoid. 

It is evident that if a plane be made to pass 
through the opposite edges of the upper and ' ^ ^ 
lower bases, the prismoid will be divided into 
two wedges, whose bases are the bases of the 



prismoid, and whose edges are L and I. The solidity of these 
wedges, and, consequently, that of the prismoid, is 

iBA(2L + r) + iJA(2Z + L)=iA(2BL + BZ+2JZ + ^'L). 
But, since M is equally distant from L and Z, we have 
2M = L + /, and 2m=B + &; 
hence 4:Mm = (L + Z) (B + 2-) = BL + B^ + &L + U. 

Substituting 4M??^ for its value in the preceding expression, 
we obtain for the solidity of the prismoid 
^A(BL + 5Z + 4M??2). 
Ex. 1. What are the contents of a log of wood, in the form 
of a rectangular prismoid, the length and breadth of one end 
being 16 inches and 12 inches, and of the other end 7 inches 
and 4: inches, the length of the log being 24 feet? 

Ans.., 16^ cubic feet. 
Ex. 2. What is the solidity of a log of hewn timber, whose 
ends are 18 inches by 15, and 14 inches by 11-|-, its length be- 
ing 18 feet ? Ans., 26fJ- cubic feet. 

Problem IX. 

To compute the excavation or emlanTcment for a raihoay. 

132. By the preceding rule may be computed the amount 
of excavation or embankment required in constructing a rail- 
road or canal. If we divide the line of the road into portions 
so small that eacli may be regarded as a straight line, and 



84 



Tkigonometet. 



suppose an equal number of transverse sections to be made, 
the excavation or embankment between two sections may be 
regarded as a prismoid, and its contents found by the pre- 
ceding rule. 

Let ABCD represent the lower surface of the supposed ex- 
cavation, which we will assume to be parallel to the horizon ; 
and let EFGH represent the upper surface of the excavation 
projected on a horizontal plane. Also, let E'A'BT', G'C'D'H' 
represent the vertical sections at 
the extremities. If we suppose ver- 
tical planes to pass through the lines 
AC, BD, the middle part of the ex- 
cavation, or that contained between 
these vertical planes, will be a rect- 
angular prismoid, of which A^B^IvI 
will be one base, and C'D^ML the 
other base. Its solidity w^ill there- 
fore be given by Art. 131. The 
parts upon each side of the middle prismoid are also halves of 
rectangular prismoids ; or, if the two parts are equal, they 
may be regarded as constituting a second prismoid, one of 
whose bases is the sum of the triangles A^E^I, BT'K ; and the 
other base is the sum of the triangles C^GT, D^H'M. There- 
fore the volume of the entire solid is equal to the product of 
one sixth of its length, by the sum of the areas of the sections 
at the two extremities, and four times the area of a parallel and 
equidistant section. 

Ex. 1. Let ABCDEFG represent the profile of a tract of land 




selected for the line of a railway ; and suppose it is required, 

by cutting and embankment, to reduce it from its present hilly 

surface to one uniform slope from the point A to the point G. 

The distance AH is 561 feet ; the distance DK is 820 feet ; 

" " HI is 858 feet ; " " KL is 825 feet ; 



ID is 825 feet; 



LG is 330 feet. 





Mensukation of Solids. 85 

The perpendicular BH is 18 ft. ; the perpendicular KE is 19 ft. ; 

« " CI is 20 ft.; " " LFis 8ft. 

The annexed figure repre- 
sents a cross section, showing 
the form of the excavation. 

The base of the cutting is to 
be 50 feet wide, the slope IJ horizontal to 1 perpendicular ; 
that is, where the depth ad is 10 feet, the width of the slope 
cd at the surface w^ill be 15 feet. 

Calculation of the jportion ABH. 
Since BH is 18 feet, the length of cd in the cross section 
will be 27 feet, and cf, the breadth at the top of the section, 
will be 10-1 feet. We accordingly find, by Art. 104, the area of 
the trapezoid forming the cross-section at BH equal to 

i5*±^^ 18 = 1386 feet. 

For the middle section the height is 9 feet, cd is 13.5 feet, 
and cf is 77 feet. The area of the cross section is therefore 
equal to 

^^ 9 = 571.5. 

The solid ABH will therefore be equal to 

(1386 + 4x571.5) -^ = 343332 cubic feet, or 

12716 cubic yards. 

Calculation of the portion BCIH. 
Since CI is 20 feet, the length of cd is 30 feet, and cf is 110 
feet. The area of the section at CI is therefore equal to 

HO+^X 20 = 1600. 

For the middle section, the height is 19 feet, cd is 28.5 feet, 
and c/'is 107 feet. The area of the cross section is therefore 
equal to 

i^x 19 = 1491.5. 

The solid BCIH will therefore be equal to 



85 



Teigonometky. 



QKQ 

(1386 + 1600 + 4 X 1491.5) -^ = 1280136 cubic feet, or 
47412.4 cubic yards. 

Calculation of the portion CID. 
The height of the middle section is 10 feet ; therefore cf is 
80 feet, and the area of the cross section is 

80 + 50 



2 



X 10 = 650. 



The solid CID will therefore be equal to 

825 
(1600 + 4 X 650) -g- r= 577500 cubic feet, or 

21388.9 cnbic yards. 
The entire amount of excavation therefore is, 
ABH = 12716 cubic yards. 
BCIH=: 47412.4 " 

GDI :^ 21388.9 
Total excavation, 81517.3 '' 



The following is a cross section, showing the form of the 
embankment. 

The top of the embankment 
is to be 50 feet wide, the slope 
2 to 1 ; that is, where the height 
ad is 10 feet, the base cd is to 
be 20 feet. 




Calcidation of the portio7i DKE. 
Since EK is 19 feet, the length of cd is 38 feet, and cf is 
126 feet. The area of the cross section at EK is therefore 
equal to 

i?^x 19 = 1672. 

For the middle section, the height is 9.5 feet; cd is there- 
fore 19 feet, and cf is 88 feet. The area of the cross section 
is therefore 

5^±^x 9.5 = 655.5 



Mensueation of Solids. 87 

The solid DKE is therefore equal to . 

890 
(1672 + 4 X 655.5) -^ = 586846.7 cubic feet, or 

21735.1 cubic yards. 

Calculation of the jyortion KEFL. 
, Since LF is 8 feet, <?(^ is 16 feet, and cf is 82 feet. The area 
of the section at LF is therefore equal to 

M, 8 = 528. 

The height of the middle section is 13.5 feet ; therefore cd 
is 27 feet, and cf is 104 feet. The area of the cross section is 
therefore equal to 

l°*±^x 13.5 = 1039.5 

The solid KEFL will therefore be equal to 

8'>5 
(1672 + 528 + 4x1039.5) -^==874225 cubic feet, or 

32378.7 cubic yards. 

Calculation of the portion LEG. 
The height of the middle section is 4 feet ; therefore cf is 
QQ feet, and the area of the cross section is equal to 

5^x4 = 232. 

The solid LEG will therefore be equal to 

(528 + 4 X 232) -^ = 80080 cubic feet, or 

2965.9 cubic yards. 
The entire amount of embankment therefore is 
DKE = 21735.1 cubic yards. 
KEFL = 32378.7 " 

LEG= 2965.9 " 

Total embankment, 57079.7 " 

Ex. 2. Compute the amount of excavation of the hill ABCD 
from the following data: 

The distance AH is 325 feet ; the perpendicular BII is 12 feet ; 
" '^ III is 672 feet; " " CI is 13 feet; 

" " ID is 534 feet. 



88 Tkigonometey. 

The base of the cutting to be 50 feet wide, and the slope IJ 
horizontal to 1 perpendicular. Ans., 33969 cubic yards. 

Peoblem X. 

133. To find the surface of a regular poly edr on. 

Rule. 

Midtij^ly the area of one of the faces hy the nuinber of 
faces ; or, Multiply the square of 07ie of the edges hy the sur- 
face of a similar solid whose edge is iinity. 

Since all the faces of a regular poljedron are equal, it is 
evident that the area of one of them, multiplied by their num- 
ber, will give the entire surface. Also, regular solids of the 
same name are similar, and similar polygons are as the squares 
of their homologous sides {Geom.^ Prop. 27, B. lY.). The fol- 
lowing table shows the surface and solidity of regular poly- 
edrons whose edge is unity. The surface is obtained by mul- 
tiplying the area of one of the faces, as given in Art. 109, by 
the number of faces. Thus tlie area of an equilateral trian- 
gle, whose side is 1, is 0.4330127. Hence the surface of a 
regular tetraedron 

= .4330127x4.= 1.7320508, 
and so on for the other solids. 

A Talle of the regidar Polyedrons whose Edges are unity. 

Names. No. of Faces. Surface. Solidity. 

Tetraedron, 4 1.7320508 0.1178513. 

Hexaedron, 6 6.0000000 1.0000000. 

Octaedron, 8 3.4641016 0.4714045. 

Dodecaedron, 12 20.6457288 7.6631189. 

Icosaedron, 20 8.6602540 2.1816950. 

Ex. 1. What is the surface of a regular octaedron whose 
edges are each 8 feet ? Ans., 221.7025 feet. 

Ex. 2. What is the surface of a regular dodecaedron whose 
edge is 12 feet ? Ans., 2972.985 feet. 

Problem XL 

134. To find the solidity of a regular polyedron. 



Mexsukation of Solids. 89 

EULE. 

Multiply the surface hy one third of the jperpendiculccr let 
fall from the centre on one of the faces / or, Midtijply the cube 
of one of the edges hy the solidity of a similar j^olyedron^ whose 
edge is unity. 

Since the faces of a regular poljedron are similar and equal, 
and the solid angles are all equal to each other, it is evident 
that the faces are all equally distant from a point in the solid 
called the centre. If planes be made to pass through the cen- 
tre and the several edges of the solid, thej will divide it into 
as many equal pyramids as it has faces. The base of each 
pyramid will be one of the faces of the polyedron ; and since 
their altitude is the perpendicular from the centre upon one of 
the faces, the solidity of the polyedron must be equal to the 
areas of all the faces, multiplied by one third of this perpen- 
dicular. 

Also, similar pyramids are to each other as the cubes of 
their hom^ologous edges {Geom.^ Prop. 17, Cor. 3, B. YIII.). 
And since two regular polyedrons of the same name may be 
divided into the same number of similar pyramids, they must 
be to each other as the cubes of their edges. 

135. The solidity of a tetraedron whose edge is unity, may 
be computed in the following manner : 

Let C — ABD be a tetraedron. From one angle, C, let fall 
a perpendicular, CE, on the opposite face ; c 

draw EF perpendicular to AD ; and join 
CF, AE. Then AEF is a right-angled tri- 
angle, in which EF, being the sine of 30°, 
is one half of AE or BE ; and therefore 
FE is one third of BF or CF. Hence the 
cosine of the angle CFE is equal to \ ; that jC b 

is, the angle of inclination of the faces of the polyedron is 
70° 31^ 4i^^ Also, in the triangle CAF, CF is the sine of 
60°, which is 0.866025. Hence, in the right-angled triangle 
CEF, knowing one side and the angles, we can compute CE, 
which is found to be 0.8164966. Whence, knowing the base 
ABD (Art. 109), we obtain the solidity of the tetraedron = 
0.1178513. 




90 Tkigonometet. 

In a somewhat similar manner may the solidities of the 
other regular poljedrons, given in Art. 133, be obtained. 

Ex. 1. What is the solidity of a regular tetraedron whose 
edges are each 24 inches ? A71S., 0.9428 feet. 

Ex. 2. What is the solidity of a regular icosaedron whose 
edges are each 20 feet ? Ans., 17453.56 feet. 

THE THREE ROUND BODIES. 
Problem I. 

136. To find the surface of a cylinder. 

EULE. 

Multiply the circumference of the lase hy the altitude for 
the convex surface. To this add the areas of the two ends 
when the entire surface is required. 

See Geometry, Prop. 1, B. X. 

Ex. 1. What is the convex surface of a cylinder whose alti- 
tude is 23 feet, and the diameter of its base 3 feet? 

Ans.^ 216.77 square feet. 

Ex. 2. What is the entire surface of a cylinder whose alti- 
tude is 18 feet, and the diameter of its base 5 feet ? 

Ans. 

Peoblem XL 

137. To find the solidity of a cylinder. 

KULE. 

Multiply the area of the lase hy the altitude. 

See Geometry, Prop. 2, B. X. 

Ex. 1. What is the solidity of a cylinder whose altitude is 
18 feet 4 inches, and the diameter of its base 2 feet 10 inches? 

Ans.., 115.5917 cubic feet. 

Ex. 2. What is the solidity of a cylinder whose altitude is 
12 feet 11 inches, and the circumference of its base 5 feet 3 
inches ? Ans.. 28.3308 cubic feet. 



Problem III. 
138. To find the surface of a cone. 



Mensuration of Solids. 91 

Rule. 

Multiply the circuwference of the hase hy half the side for 
the convex surface / to which add the area of the lase V)hen the 
entire surface is reqidred. 

See Geometry, Prop. 3, B. X. 

Ex. 1. What is the entire surface of a cone whose side is 10 
feet, and the diameter of its base 2 feet 3 inches ? 

Ans.^ 39.319 square feet. 

Ex. 2. What is the entire surface of a cone whose side is 15 
feet, and the circumference of its base S feet? 

Ans.^ 65.093 square feet. 

Pkoblem IY. 

139. To find the solidity of a cone. 

Rule. 
Multiply the area of the lase hy one third of the altitude. 
See Geometry, Prop. 5, B. X. 

Ex. 1. Wliat is the solidity of a cone whose altitude is 12 
feet, and the diameter of its base 2J feet? 

Ans.^ 19.635 cubic feet. 
Ex. 2. What is the solidity of a cone whose altitude is 25 
feet, and the circumference of its base 6 feet 9 inches? 

Ans. 

Peoblem Y. 

140. To find the surface of a frustum of a cone. 

Rule. 

Midtiply half the side hy the sum of the circumferences of 
the tiuo hasesfor the convex surface ; to this add the areas of 
the tioo hases when the entire surface is required. 

See Geometry, Prop. 4, B. X. 

Ex. 1. Wliat is the entire surface of a frustum of a cone, the 
diameters of whose bases are 9 feet and 5 feet, and whose side 
is 16 feet 9 inches? Ans.^ 451.6036 square feet. 

Ex. 2. What is the convex surface of a frustum of a cone 



92 Trigonometry. 

whose side is 10 feet, and the circumferences of its bases 6 feet 
and 4 feet ? Ans., 50 square feet. 

Problem VI. 

141. To find the solidity of a frustum of a cone. 

Rule. 

Add together the areas of the two hases^ and a mean proj)or- 
tional letween them^ and inidti^ly the sum hy one third of the 
altitu.de. 

See Geometry, Prop. 6, B. X. 

If we put R and r for the radii of the two bases, then ttP^ 
will represent the area of one base, irr"^ the area of the other, 
and ttP^ the mean proportional between them. Hence, if we 
represent the height of the frustum by A, its solidity will be 

Ex. 1. What is the solidity of a frustum of a cone whose al- 
titude is 20 feet, the diameter of the greater end 5 feet, and 
that of the less end 2 feet 6 inches ? 

Ans., 229.074 cubic feet. 

Ex. 2. The length of a mast is 60 feet, its diameter at the 
greater end is 20 inches, and at the less end 12 inches : what 
is its solidit}^? Ans.^ 85.521 cubic feet. 

Problem YIL 

142. Tofimd the surface of a sphere. 

Rule. 

Multiply the diameter hy the circumference of a great circle; 
or. Multiply the square of the diameter hy 3.14159. 

See Geometry, Prop. 7, B. X. 

Ex. 1. Required the surface of the earth, its diameter being 
7912 miles. Ans., 196,662,896 square miles. 

Ex. 2. Required the surface of the moon, its circumference 
being 6786 miles. An^. 

Problem YIII. 
, 143. Tofimd the solidity of a sphere. 



Mexsueation of Solids. 



93 



Rule. 

Midtij^ly the surface hy one third of the radius ; or, 2iid- 
tiply the cube of the diameter hy \-n ; that is, hy 0.5236. 

See Geometry, Prop. 8, B. X. 

TThere great accuracy is required, the value of ^tt must be 
taken to more tliau four decimal places. Its value, correct to 
ten decimal places, is .52359,87756. 

Ex. 1. What is the solidity of the earth, if it be a sphere 
7912 miles in diameter ? 

Ans., 259,332,805,350 cubic miles. 

Ex. 2. If the diameter of the moon be 2160 miles, what is 
its solidity? Aiis. 

Peoblem IX. 
144. To find the surface of a spherical zone. 

EULE. 

Multijfly the altitude of the zone ly the circumference of a 
great circle of the sphere. 

See Geometry, Prop. 7, Cor. 1, B. X. 

Ex. 1. If the diameter of the earth be 7912 miles, what is 
the surface of the torrid zone, extending 23° 27' 36^^ on each 
side of the equator ? Ans.., 78,293,218 square miles. 

Let PEP'Q represent a meridian of the earth; EQ the 
equator ; P, P' the poles ; AB one of the ^ p ^ 

tropics, and GH one of the polar circles. 
Then PK will represent the height of one 
of the frigid zones, KD the height of one 
of the temperate zones, and CD half the 
height of the torrid zone. 

Each of the angles ACE, CAD, and 
GCK is equal to 23° 27^ 36^^ 

In the right-angled triangle ACD, 

CD=zACsin. CAD. 

Also, in the right-angled triangle CGK, 
CK=:CGcos. GCK. 

Then PK = PC-KC. 



./\ 


D \t. 


,/~-A 


n 


\ 


° ] 


V 


J 



94: Trigonometey. 

Where great accuracy is required, the sine and cosine of 23° 
27^ 36^^ must be taken to more than six decimal places. The 
following values are correct to ten decimal places : 
Natural sine of 23° 27' 36^' = .39810,87431. 
" cosine of 23° 27' 36'' ==. 91733,82302. 
Ex. 2. If the polar circle extends 23° 27' 36" from the pole, 
find the convex surface of either frigid zone. 

A7is.^ 8,128,252 square miles. 
Ex. 3. On the same supposition, find the surface of each of 
the temperate zones. Ans., 51,056,587 square miles. 

Peoblem X. 

145. To find the solidity of a spherical segment with one 
hase. 

Rule. 
Multiply half the height of the segment ty the area of the 
lase, and the cube of the height hy .5236, and add the two 
products. 
See Geometry, Prop. 9, B. X. 

Ex. 1. What is the solidity of either frigid zone, supposing 
the earth to be 7912 miles in diameter, the polar circles ex- 
tending 23° 27' 36" from the poles? 

Ans., 1,292,390,176 cubic miles. 

146. The solidity of a spherical segment of two bases is 
the difference between two spherical segments, each having a 
single base. 

Ex. 2. On the same supposition as in Ex. 1, find the solidity 
of either temperate zone, 

Ans., 55,032,766,543 cubic miles. 
Ex. 3. Find the solidity of the torrid zone. 

Ans., 146,682,491,911 cubic miles. 

Peoblem XI. 

147. To find the area of a spherical triangle. 

Rule. 
Compute the surface of the guadrantal triangle, or one 
eighth of the surface of the sp^here. From the sum of the three 



Mensueation of Solids. 95 

angles siibtract two right angles ; divide the remainder ly 90, 
and multijphj the quotient hy the guadrantal triangle. 

See Geometiy, Prop. 20, B. IX. 

Ex. 1. What is the area of a triangle on a sphere whose di- 
ameter is 10 feet, if the angles are 55°, 60°, and 85° ? 

Ans.^ 8.7266 square feet. 

Ex. 2. If the angles of a spherical triangle measured on the 
surface of the earth are 78° ^ 10'^ 59° 50' 54:'^ and 42° 5' ZT', 
what is the area of the triangle, supposing the earth a sphere, 
of which the diameter is 7912 miles? 

Ans.^ 3110.791: square miles. 

If the excess of the angles above two right angles is ex- 
pressed in seconds, we must divide it by 90 degrees also ex- 
pressed in seconds ; that is, bj 324,000. 

Problem XII. 
148. To find the area of a sjoherical polygon. 

KULE. 

Comjnde the surface of the quadrantal triangle. From the 
sum of all the angles suhtixict the product of tivo right angles 
hy the numher of sides less two; divide the remainder hy 90, 
and multiply the quotient hy the quadrantal triangle. 

See Geometry, Prop. 21, B. IX. 

Ex. 1. What is the area of a spherical polygon of 5 sides on 
a sphere whose diameter is 10 feet, supposing the sura of the 
angles to be 640 degrees ? Ans.^ 43.633 square feet. 

62° 33' 13'' ; 



Ex. 2. The angles of a spherical 
polygon, measured on the surface of 
the earth, are 



135° 8' 26"; 
149° 16' 9" ; 
111° 45' 8"; 
105° 59' 7"; 
155° 19' 12": 



Required the area of the polygon. 

Ans.^ 5690.477 square miles. 

149. Examples for Practice. 

Prob. 1. The base of a triangle is 20 feet, and its altitude 
18 feet. It is required to draw a line parallel to the base so as 



96 Teigonometey. 

to cut off a trapezoid containing 80 square feet. What is the 
length of the line of section, and its distance from the base of 
the triangle ? 

Ans. Length 14.907 feet; distance from base 4.584 feet. 

PtoI). 2. The base of a triangle is 20 feet, one angle at the 
base is 63° 26', and the other angle at the base is 56° 19'. It 
is required to draw a line parallel to the base, so as to cut off a 
trapezoid containing 109 square feet. What is the length of 
the line of section, and its distance from the base of the tri- 
angle % 

Ans. Length 12.070 feet ; distance from base 6.797 feet. 

Prob. 3. In a perpendicular section of a ditch, the breadth 
at the top is 26 feet, the slopes of the sides are each 45°, and 
the area 140 square feet. Kequired the breadth at bottom and 
the depth of the ditch. 

Ans. Breadth 10.77 feet ; depth 7.615 feet. 

Prob. 4. The altitude of a trapezoid is 23 feet ; the two par- 
allel sides are 76 and 36 feet ; it is required to draw a line par- 
allel to the parallel sides, so as to cut off from the smaller end 
of the trapezoid a part containing 560 square feet. What is 
the length of the line of section, and its distance from the 
shorter of the two parallel sides ? 

A71S. Length 56.954 feet ; distance 12.048 feet. 

Prob. 5. From the greater end of a trapezoidal field whose 
parallel ends and breadth measure 12, 8, and lOJ chains re- 
spectively, it is required to cut off an area of six acres by a 
fence parallel to the parallel sides of the field. What is the 
length of the fence, and its distance from the greater side. 

Ans. Length of fence 9.914 chains ; distance from greater 
side 5.476 chains. 

Prob. 6. There are three circles whose radii are 20, 28, and 
29 inches respectively. Required the radius of a fourth circle, 
whose area is equal to the sura of the areas of the other three. 

Ans. 45 inches. 

Prob. 7. In constructing a railroad, the pathway of which is 
24 feet broad, it is necessary to make a cutting 40 feet in 
depth ; what must be the breadth of the cutting at top, suppos- 
ing the slopes of the sides to be 65° ? Ans. 61.305 feet. 



Mensueation of Solids. 97 

pToh. 8. The sides of a quadrilateral field are 690 yards, 467 
yards, 359 yards, and 428 yards ; also, the angle contained be- 
tween the first and second sides is 57° 30', and the angle be- 
tween the third and fourth sides 96° 42^ Required the area 
of the field. Ans., 212184 square yards. 

Proh. 9. There are two regular pentagons, one inscribed in 
a circle, and. the other described about it; and the difference 
of the areas of the pentagons is 100 square inches. Required 
the radius of the circle. Ans., 8.926 inches. 

Proh. 10. What is the length of a chord cutting off one 
third part of a circle, whose diameter is 289 feet. 

Solution. Let (2 = angle ACD, fig. p. 74. 

AE = r sin. a. 
CE^?" COS. a. 
circle = 77 r", Art. 115. 

sector ACD = ^ - r\ Art. 116. 



triangle AEC: 
semi-segment AED: 


=4 


r'' sin. 


a COS. 


a. 




a 
360 


^'^'-^r 


sin 


. a cos 


• «=i 


TT 


t\ 



180 . 
a — Sin. a cos. (2 + 60. 

TT 

a = 28.648 sin. 2a + 60, Art. 89. 
This equation is readily solved by the method of approxima- 
tion. Algebra, Art. 470. 
' « = 74°.637. 

AE = 139.33 feet. Ans. 
Proh. 11. The area of a triangle is 1012; the length of the 
side a is to that of Z> as 4 to 3, and c is to Z> as 3 to 2. Required 
the length of the sides. 

Ans., « = 52.470, ^==39.353, ^ = 59.029. 
Prob. 12. The area of a triangle is 144, the base is 24, and 
one of the angles at the base is 30°. Required the other sides 
of the triangle. Am., 24 and 12.4233. 

Prol), 13. Seven men bought a grinding-stone of 60 inches 

G 



98 Trigonometey. 

<iiaineter, eacli paying one seventh part of the expense. What 
part of the diameter must each grind down for his share? 

Ans. The 1st, 4.4508 inches; 2d, 4.8400 inches; 3d, 5.3535 
inches; 4th, 6.0765 inches ; 5th, 7.20T9 inches; 6th, 
9.3935 inches ; 7th, 22.6778 inches. 

Prob. 14. The area of an equilateral triangle is 17 square 
feet and 83 square inches. What is the length of each side? 

A71S. 76.45 inches. 

Py'oh. 15. The parallel sides of a trapezoid are 20 and 12 
feet, and the other sides are 15 and lY feet. Kequired the 
area of the trapezoid. Ans.^ 240 square feet. 

Prob. 16. How many square yards of canvas are required 
to make a conical tent which is 20 feet in diameter and 12 feet 
high ? Ans.^ 54.526 square yards. 

Prob. 17. The circumference of an hexagonal pillar is 7 feet, 
and the height 11 feet 2 inches. Required the solid contents 
of the pillar. Ans..^ 39.488 cubic feet. 

Prob, 18. The base of the great pyramid of Egypt is a square 
whose side measures 746 feet, and the altitude of the pyramid 
is 450 feet. Required the volume of the pyramid. 

Ans., 83,477,400 cubic feet. 

Prob. 19. A side of the base of a frustum of a square pyra- 
mid is 25 inches, a side of the top is 9 inches, and the height 
is 20 feet. Required the volume of the frustum. 

J.715., 43.102 cubic feet. 

Prob. 20. Three persons, having bought a sugar-loaf, would 
divide it equally among them by sections parallel to the base. 
It is required to :find the altitude of each person's share, sup- 
posing the loaf to be a cone whose height is 20 inches. 

Ans., 13.8672, 3.6044, and 2.5284 inches, 

Prob. 21. If a cubical foot of brass were to be drawn into 
wire of one tliirtieth of an inch in diameter, it is required to 
determine the length of the said wire, allowing no loss in the 
metaL Ans., 55003.94 yards; or 31 miles 443.94 yards. 

Prob. 22. How high above the surface of the earth must a 
person be raised to see one third of its surface? 

Ans. The height of its diameter. 

Prob. 23. If a heavy sphere, whose diameter is 4 inches, be 
let fall into a conical glass full of water, whose diameter is 5, 



Mensuration of Solids. 



99 



and altitude 6 inches, it is required to determine liow much 
water will run over. Ans.y 26.272 cubic inches. 

Solutio7i. 



AF 




6.6 

:EC: CD .= 5.2 



2.5"-' = 

AD 

CF=zrD-CD = 0.8 
1F = 2.8; FG = 1.2; vFff = 2.8xl.2. 
The immersed segment 

= 7rx^xFff+^Fr 

2 6 



=.^FI(3Fff + Fr) 

=:|-x 2.8 (10.08 + 2.8^) 



X 50.176 = 26.27209 



jProh. 24. The capacity of a cylinder is a cubic feet, and its 
convex surface is h square feet. Kequired the dimensions of 
the cylinder. 

2a 



Ans. Eadius of base = -t", and altitude 



7 , ana aitituae^- — . 

Prob. 25. A triangular pyramid, the sides of whose base are 
13, M, and 15 inches respectively, and whose altitude is 16 
inches, is cut, at the distance of 2 inches from the vertex, by a 
plane parallel to the base. Required the volume of the frus- 
tum of the pyramid. Ans.^ 447.125 cubic inches. 

Prob. 26. The altitude of a cone is 10 inches, and the radius 
of its base is 5 inches. At what distance from the base must 
a plane pass parallel to the base, so as to cut off a frustum 
wljose capacity is 20 cubic inches? Ans., 0.2614 inches. 



CHAPTER lY. 

SURVEYING. 

150. The term Surveying includes the measurement of 
heights and distances, the determination of the area of portions 
of the earth's surface, and their delineation upon paper. 

Since the earth is spherical, its surfade is not a plane surface, 
and if large portions of the earth are to be measured, the cur- 
vature must be taken into account; but in ordinary surveying, 
the portions of the earth are supposed to be so small that the 
curvature may be neglected. The parts surveyed are there- 
fore regarded as plane figures. 

151. If a plummet be freely suspended by a line, and al- 
lowed to come to a state of rest, this line is called a vertical 
line. 

Every plane passing through a vertical line is a vertical 
plane. 

A line perpendicular to a vertical line is a horizontal line. 

A plane perpendicular to a vertical line is a horizontal jplane. 

A vertical angle is one the plane of whose sides is vertical. 

A horizontal angle is one the plane of whose sides is hori- 
zontal. 

An angle of elevation is a vertical angle having one side 
horizontal and the other an ascending ^ D 

line, as the angle BAD. 

An angle of depression is a vertical 
angle having one side horizontal and 
the other a descending line, as the an- 
gle CDA. 

152. When distances are to be found A b 
by trigonometrical computation, it is necessary to measure at 
least one line upon the ground, and also as many angles as 
may be required to render three parts of every triangle 
known. 




Surveying. 101 

In the measurement of lines, the unit commonly employed 
by surveyors is a chain four rods or sixty-six feet in length, 
called Gunter's Chain, from the name of the inventor. This 
cliain is divided into 100 links. Sometimes a half chain is 
used, containing 60 links. 

Hence, 1 chain = 100 links =Q^ feet; 
1 rod = 25 links rz.l6ifeet; 
1 link =7.92 inches^ f of a foot nearly. 

153. To measure a horizontal line. 

To mark the termination of the chain in measuring, ten iron 
pins should be provided, about a foot in length. 

Let the person who is to go foremost in carrying the chain, 
and who is called the leader, take one end of the chain and 
the ten pins ; and let another person take the other end of the 
cliain, and hold it at the beginning of the line to be measured. 
When the leader has advanced until the chain is stretched 
tight, he must set down one pin at the end of the chain, the 
other person taking care that the chain is in the direction of 
the line to be measured. Then measure a second chain in the 
same manner, and so on until all the marking pins are ex- 
hausted. A record should then be made that ten chains have 
been measured, after which the marking pins should be re- 
turned to the leader, and the measurement continued as before, 
until the whole line has been passed over. 

It is generally agreed to refer all surfaces to a horizontal 
plane. Hence, when an inclined surface, like the side of a 
hill, is to be measured, the chain should be maintained in a 
horizontal position. For this purpose, in ascending a hill, the 
hind end of the chain should be raised from the ground until 
it is on a level with the fore end, and should be held vertically 
over the termination of the preceding chain. In descending 
a hill, the fore end of the chain should be raised in the same 
manner. 



INSTRUMENTS FOR MEASURING ANGLES. 

In measuring angles, some instrument is used which con- 
tains a portion of a graduated circle divided into degrees and 
minutes. These instruments ma}^ be adapted to measuring 



102 



Teigonometet. 



either horizontal or vertical angles. The instrument most fre- 
quently employed for measuring horizontal angles is called 

The Sueveyok's Compass. 
154. The principal parts of this instrument are a compass- 
box, a magnetic needle, two sights, and a stand for its support. 
The compass-box, ABO, is circular, generally about six inches 
in diameter, and at its centre is a small pin on which the mag- 
netic needle is balanced. The circumference of the box is di- 
vided into degrees, and sometimes to half degrees ; and the 
degrees are numbered from the extremities of a diameter both 
ways to 90°. The sights, DE, FG, are placed at right angles 




to the plane of the graduated circle, and in each of them there 
is a large and small aperture for convenience of observation. 
The instrument, when used, is mounted on a tripod, or a single 
staff pointed with iron at the bottom, so that it may be firmly 
placed in the ground. 

Sometimes two spirit levels, H and K, are attached, to indi- 
cate when the plane of the graduated circle is brought into a 
horizontal position. 

155. "When the magnetic needle is supported so as to turn 
freely, and is allowed to come to a state of rest, the direction 
it assumes is called the magnetic meridian^ one end of the 
needle indicating the north point and the other the south. 

A horizontal line perpendicular to a meridian is an east and 
west line. 



Surveying. 



103 



N 



All the meridians passing throngli a survey of moderate ex- 
tent are considered as straight lines parallel to each other. 

The hearing or course of a line is the angle which it makes 
with a meridian passing through one end ; and it is reckoned 
from the north or south point of the horizon, toward the east 
or west. 

Thus, if ]N"S represent a meridian, and the angle NAB is 40°, 
then the hearing of AB from the point A is 
40° to the west of north, and is written ]^. 40° 
W., and read north forty degrees west. 

The reverse hearing of a line is the bearing 
taken from the other end of the line. 

The forward bearing and reverse bearing 
of a line are equal angles, but lie between di- 
rectly opposite points. Thus, if the bearing 
of AB from A is N. 40° W., the bearing of the 
same line from B is S. 40° E. 

156. For measuring vertical angles, the instrument com- 
monly used is 

A Quadrant. 

It consists of a quarter of a circle, usually made of brass, 

and its limb, AB, is divided into c>. j^ 

degrees and minutes, numbered 
from A up to 90°. It is furnished 
either with a pair of plain sights 
or with a telescope, CD, which is 
to be directed toward the object 
observed. A plumb line, CE, is 
suspended from the centre of the 
quadrant, and indicates when the 
radius CB is brought into a ver- 
tical position. 

To measure the angle of elevation, for example, of the top 
of a tower, point the telescope, CD, toward the tower, keeping 
the radius, CB, in a vertical position by means of the plumb 
line, CE. Move the telescope until the given object is seen in 
the middle of the field of view. The centre of the field is in- 
dicated by two wires placed in the focus of the object-glass of 




104: 



Trigonometry, 



the telescope, one wire being vertical and the other horizontal. 
When the horizontal wire is made to coincide with the summit 
of the tower, the angle of elevation is shown upon the arc AB 
by means of an index which moves with the telescope. 

As the arc is not commonly divided into parts smaller than 
half degrees, when great accuracy is required, some contrivance 
is needed for obtaining smaller fractions of a degree. This is 
usually effected by a vernier. 

157. A Vernier is a scale of small extent, graduated in 
such a manner that, being moved by the side of a fixed scale, 
w^e are enabled to measure minute portions of this scale. The 
length of this movable scale is ^equal to a certain number of 
parts of that to be subdivided, but it is divided into parts one 
more or one less than those of the primary scale taken for the 
length of the vernier. Thus, if we wish to measure hundredths 
of an inch, as in the case of a barometer, we first divide an 
inch into 10 equal parts. We then construct a vernier equal 
in length to 11 of these divisions, but divide it into 10 equal 
parts, by w^hieh means each division on the vernier is yL 
longer than a division of the primary scale. 

Thus, let AB be the upper end of a barometer tube, the mer- 
cury standing at the point C ; the scale is ^ 
divided into inches and tenths of an inch, 
and the middle piece, numbered from 1 
to 9, is the vernier that slides up and 
down, having 10 of its divisions equal to 
11 divisions of the scale, that is, to -fj- 
of an inch. Therefore, each division of 
the vernier is -J-gL- of an inch ; or one 
division of the vernier exceeds one divi- 
sion of the scale by -j^-^ of an inch. 
Now, as the sixth division of the vernier 
(in the figure) coincides with a division 
of the scale, the fifth division of the ver- 
nier will stand -^^ of an inch above the nearest division of 
the scale; the fourth division ^-J-q of an inch, and the top 
of the vernier will be ^fo- ^^ ^^^ i"ch above the next lower 
division of the scale ; i. e.^ the top of the vernier coincides with 
29.66 inches upon the scale. In practice, therefore, we ob- 




SUKYETING. 105 

serve what division of the vernier coincides with a division of 
the scale ; this will show the hundredths of an inch to be added 
to the tenths next below the vernier at the top. 

A similar contrivance is applied to graduated circles, to ob- 
tain the value of an arc with greater accuracy. If a circle is 
graduated to half degrees, or 30^, and we wish to measure sin- 
gle minutes by the vernier, we take an arc equal to 31 divi- 
sions upon the limb, and divide it into 30 equal parts. Then 
each division of the vernier will be equal to |-i- of a degree, 
while each division of the scale is -|-2- of a degree. That is, 
each space on the vernier exceeds one on the limb by V. 

In order, therefore, to read an angle for any position of the 
vernier, we pass along the vernier until a line is found coin- 
ciding with a line of the limb. The number of this line from 
the zero point indicates the minutes which are to be added to 
the degrees and half degrees taken from the graduated circle. 
Sometimes a vernier is attached to the common surveyor's 
compass. 

158. An instrument in common use for measuring both 
horizontal and vertical angles is 

The Theodolite. 

The theodolite has two circular brass plates, C and D (see fig. 
next page), the former of which is called the vernier plate, and 
the latter the graduated lirab. Both have a horizontal motion 
about the vertical axis, E. This axis consists of two parts, one 
external, and the other internal ; the former secured to the 
graduated limb, D, and the latter to the vernier plate, C, so that 
the vernier plate turns freely upon the lower. The edge of the 
lower plate is divided into degrees and half degrees, and this 
is subdivided by a vernier attached to the upper plate into 
single minutes. The degrees are numbered from to 360. 

The parallel plates, A and B, are held together by a ball 
which rests in a socket. Four screws, three of which, a, a, a, 
are shown in the figure, turn in sockets fixed to the lower plate, 
while their heads press against the under side of the upper 
plate, by which means the instrument is levelled for observa- 
tion. The whole rests upon a tripod, which is firmly attached 
to the body of the instrument. 



106 



Teigonometrt. 




To tlie vernier plate, two spirit-levels, <?, <?, are attached at 
right angles to each other, to determine when the graduated 
limb is horizontal. A compass, also, is placed at F. Two 
frames, one of which is seen at N, support the pivots of the 
horizontal axis of the vertical semicircle KL, on which the tel- 
escope, GH, is placed. One side of the vertical arc is divided 
into degrees and half degrees, and it is divided into single min- 
utes by the aid of its vernier. The graduation commences at 
the middle of the are, and reads both ways to 90°. Under and 
parallel to the telescope is a spirit-level, M, to show when the 
telescope is brought to a horizontal position. To enable us to 
direct the telescope npon an object with precision, two lines, 
called wires, are fixed at right angles to each other in the focus 
of the telescope. 

To measure a Horizontal Angle with the Theodolite. 
159. Place the instrument exactly over the station from 
which the angle is to be m.easured ; then level the instrument 
by means of the screws, (2, a^ bringing the telescope over each 
pair alternately until the two spirit-levels on the vernier plate 
retain their position, while the instrument is turned entirely 
round upon its axis. Direct the telescope to one of \\\(i objects 



SUEYEYIXG. 107 

to be observed, moving it until the cross-wires and object co- 
incide. JSTow read off the degrees upon the graduated limb, 
and the minutes indicated bv the vernier. iS'ext, release the 
upper plate (leaving the graduated limb undisturbed), and 
move it round until the telescope is directed to the second ob- 
ject, and make the cross-vrircs bisect this object, as was done 
by the first. Again, read off the vernier ; the difference be- 
tween this and the former reading will be the angle required. 
The magnetic bearing of an object is determined by simply 
reading the angle pointed out by the compass-needle when 
the object is bisected. 

To measure an Angle of Elevation loith the Theodolite. 

160. Direct the telescope toward the given object so that 
it may be bisected by the horizontal wire, and then read off the 
arc upon the vertical semicircle. After observing the object 
with the telescope in its natural position, it is well to revolve 
the telescope in its supports until the level comes uppermost, 
and repeat the observation. The mean of the two measures 
may be taken as the angle of elevation. 

By the aid of the instruments now described, we may de- 
termine the distance of an inaccessible object, and its height 
above the surface of the earth. 

HEIGHTS AXD DISTAXCES. 
Problem I. 

161. To determine the height of a vertical object situated 
on a horizontal plane. 

Measure from the object to any convenient distance in a 
straight line, and then take the angle of elevation subtended 
by the object. 

If we measure the distance DE, and 
the angle of elevation CDE, there will 
be given, in the right-angled triangle 
CDE, the base and the angles, to find 
the perpendicular CE (Art. 53). To 
this we must add the hei^'ht of the in- 
strument, to obtain the entire height 
of tlie object above the plane AB. 



/ 

/ 

/ 
/ 

/ 


1 

1 


1 


1 



/ 



1 



|-T 



108 Teigonometey. 

Ex. 1. Having measured AB equal to 100 feet from the 
bottom of a tower on a horizontal C 4 

plane, I found the angle of elevation, ,, 

ODE, of the top to be 47° 30^, the / ^ 

centre of the quadrant being five 

feet above the ground. What is the / 

heiglit of the tower ? 

CE = DE tan. CDE--= 109.13. 

To which add five feet, and we obtain ^^ 

the height of the tower, 114.13 feet, a b 

Ex. 2. From the edge of a ditch 18 feet wide, surround- 
ing a fort, the angle of elevation of the wall was found to 
be 62° 40^ Required tl^e height of the wall, and the length 
of a ladder necessary to reach from my station to the top 
of it. 

A71S. The height is 34.82 feet. Length of ladder, 39.20 feet. 

Problem II. 

162. To find the distance of a vertical olject wJiose height 
is known. 

Measure the angle of elevation, and we shall have given the 
angles and perpendicular of a right-angled triangle to find the 
base (Art. 53). 

Ex. 1. The angle of elevation of the top of a tower whose 
height was known to be 143 feet, was _ C 

found to be 35°. What was its dis- 
tance ? 

Here we have given the angles of the 
triangle ABC, and the side CB, to find 
AB. Ans., 204.22 feet. 

If the observer were stationed at the 
top of the tower BC, he might find the -^ 
length of the base AB by measuring the angle of depression 
DCA, which is equal to BAG. 

Ex. 2. From the top of a ship's mast, which was 80 feet 
above the water, the angle of depression of another ship's hull 
was found to be 20°. What was its distance ? 

^71.?., 219.80 feet. 




SUEVETING. 



109 




Peoblem III. 

163. To find the height of a vertical object standing on an 
inclined jplane. 

Measure the distance from the object to any convenient sta- 
tion, and observe the angles which the base-hne makes with 
lines drawn from its two ends to the top of the object. 

If we measure the base-line AB, and the two angles ABC, 
BAG, then, in the triangle ABC, 
we shall have given one side and 
the angles to find BC. 

Ex. 1. Wanting to know the 
height of a tower standing on an 
inclined plane, BD, I measured 
from the bottom of the tower a 
distance, AB, equal to 165 feet; 
also the angle ABC, equal to 
107° 18^, and the angle BAC, 
equal to 33° 35'. Keqnired the 
height of the object. 

sin. ACB : sin. BAC ; : AB : BC = 144.66 feet. 

The height, BC, may also be found by measuring the dis- 
tances BA, AD, and taking the angles BAC, BDC. The dif- 
ference between the angles BAC and BDC will be the angle 
ACD. There will then be given, in the triangle Dx\C, one 
side and all the angles to find AC ; after which we shall have, 
in the triangle ABC, two sides and the included angle to find 
BC. 

Ex. 2. A tower standing on the top of a declivity, I meas- 
ured 75 feet from its base, and then took the angle BAC, 47° 
50' ; going on in the same direction 40 feet farther, I took the 
angle BDC, 38° 30'. What was the height of the tower? 

Ans.^ 117.21 feet. 

Peoblem IV. 

164. To find the distance of an inaccessihle olject. 
Measure a horizontal base-line, and also the angles between 

this line and lines drawn from each station to the object. Let 
C be the object inaccessible from A and B. Then, if the dis- 



no 



Teigonometet. 




tance between the stations A and B be measured, as also the 
angles at A and B, there will be given, in 
the triangle ABC, the side AB and the an- 
gles, to find AC and BC, the distances of the 
object from the two stations. 

Ex. L Being on the side of a river, and 
w^anting to know the distance to a house 
which stood on the other side, I measured A B 

400 yards in a right line by the side of the river, and found 
that the two angles at the ends of this line, formed by the 
other end and the house, were 73° 15^ and 68° 2^ What was 
the distance between each station and the house? 

Then 
sin. A : BC=:612.38; 
sin.B : AC =.593.09. 

Ex. 2. Two ships of war, wishing to ascertain their distance 
from a fort, sail from each other a distance of half a mile, w^hen 
they find that the angles formed betw^een a line from one to 
the other, and from each to the fort, are 85° 15^ and 83° 45^ 
What are tlie respective distances from the fort? 

Ans., 4584.52 and 4596.10 yards. 



The ande C is found to be 38° 43^ 



sin. C : AB : : 



Problem Y. 

165, To find the distance letioeen two oljects sejparated hy 
an iinpassdble harrier. 

Measure the distance from any convenient station to each 
of the objects, and the angle included between those lines. 

If \\Q wish to know the distance between the places C and 
B, both of which are accessible, but sep- 
arated from each other by water, we may 
measure tlie lines AC and AB, and also 
the angle A. We shall then have given 
two sides of a triangle and the included 
angle to find the third side. 

Ex. 1. The passage between the two 
objects C and B being obstructed, I measured from A to C 735 
rods, and from A to B 840 rods ; also, the angle A, equal to 
55° 40^ What is the distance of the places c'and B? 

Ans., 741.21 rods. 





w:0^' 



SUEVEYIXG. Ill 

Ex. 2. In order to find the distance between two objects, C 
and B, wbicli could not be directly measured, I measured 
from C to A 652 yards, and from B to A 756 yards ; also, the 
angle K equal to 142^ 25'. What is the distance between the 
objects C and B ? Ans. 

Peoelem W. 

166. To find the height of an inaccessible object above a 
horizontal plane. 

First Method. — Take two stations in a vertical plane pass- 
ing through the top of the object ; measure the distance be- 
tween the stations and the angle of elevation at each. 

If we measure the base AB, and the angles DAC, DBC, 
then, since GBA is the sup- 
plement of DBC, we shall 
have, in the triangle ABC, 
one side and all the angles 
to find BC. Then, in the 
right-angled triangle DBC, we shall have the hypotheuuse and 
the ano^les to find DC. 

Ex. 1. What is the perpendicular height of a hill whose an- 
gle of elevation, taken at the bottom of it, was 46°; and 100 
yards farther oH, on a level with the bottom of it, the angle 
was 31° ? Ans., 11:3.1^: yards. 

Ex. 2. The angle of elevation of a spire I found to be 5S°, 
and going 100 yards directly from it, found the angle to be 
only 32°. What is the height of the spire, supposing the in- 
strument to have been five feet above the ground at eacli ob- 
servation ? Ans., lOl.lS yards. 

167. Second Method. — Measure any convenient base-line, 
also the angles between this base and lines drawn from each 
of its extremities to the foot of the object, and the angle of 
elevation at one of the stations. 

Let DC be the given object. If we measure the horizontal 
base-line AB, and the angles CAB, CBA, we can compute the 
distance BC. Also, if we observe the angle of elevation CBD, 
we shall have given, in the right-angled triangle BCD, the 
base and angles to find the perpendicular. 



112 



Tkigonometet. 




Ex. 1. Being on one side of a river, and wanting to know 
the height of a spire on the other 
side, I measured 500 j^ards, AB, along 
the side of the river, and found the 
angle ABC=:74° U\ and BAG = 49° 
23^ ; also, the angle of elevation CBD 
= 11° 15'. Eequired the height of 
the spire. Ans., 271.97 feet. 

Ex. 2. To find the height of an in- 
accessible castle, I measured a line of 73 yards, and at each 
end of it took the angle of position of the object and the other 
end, and found the one to be 90°, and the other 61° 45' ; also, 
the elevation of the castle from the latter station, 10° 35'. 
Required the height of the castle. Ans., 86.45 feet. 

Pkoblem YII. 

168. To Ji7id the distance between two inaccessible ob- 
jects. 

Measure any convenient base-line, and the angles between 
this base and lines drawn from each of its extremities to each 
of the objects. 

Let C and D be the two inaccessible objects. If we meas- 
ure a base-line, AB, and the an- 
gles DAB, DBA, CAB, CBA, 
then, in the triangle DAB, we 
shall have given the side AB 
and all the angles to find BD ; 



also, in the triangle ABC, we 

shall have one side and all the 

angles to find BC ; and then, in 

the triangle BCD, we shall have two sides, BD, BC, with the 

included angle, to find DC. 

Ex. 1. Wanting to know the distance between a house and 
a mill, w^hich were separated from me by a river, I measured 
a base-line, AB, 300 yards, and found the angle CAB = 58° 
20', CAD = 37°, ABD = 53° 30', DBC = 45° 15'. What is the 
distance of the house from the mill? A7is.^ 479.80 yards. 

Ex. 2. Wanting to know the distance between two inaccessi- 
ble objects, C and D, I measured a base-line, AB, 28.76 rods, 






SUKYETING. 








113 


and found the 
and DEC = 76° 


angle CAB = 33°, CAD::. 
, What is the distance fr 


om 


', DBA = 
CtoD? 


;59° 


45^ 



Ans., 97.696 rods. 



THE DETERMINATION OF AREAS. 

169. The a7'ea or content of a tract of land is the horizon- 
tal surface included within its boundaries. 

When the surface of the ground is broken and uneven, it is 
very difficult to ascertain exactly its actual surface. Hence it 
has been agreed to refer every surface to a horizontal plane; 
and for this reason, in measuring the boundary lines, it is nec- 
essary to reduce them all to horizontal lines. 

The measuring unit of surfaces chiefly employed by survey- 
ors is the acre^ or ten square chains. 

One quarter of an acre is called a rood. 

Since a chain is four rods in length, a square chain contains 
sixteen square rods; and an acre, or ten square chains, con- 
tains 160 square rods. Square rods are called jperclies. The 
area of a field is usually expressed in acres, roods, and perches, 
designated by the letters A., K., P. 

When the lengths of the bounding lines of a field are given 
in chains and links, the area is obtained in square chains and 
square links. Now, since a link is y^ of a chain, a square 
link will be tq-o'^td^ ^^ ^ square chain ; that is, ^(^^^q of a 
chain. Hence we have the following 

Table. 
1 square chain = 10,000 square links. 
1 acre = 10 square chains = 100,000 square links. 
1 acre =4 roods = 160 perches. 

If, then, the linear dimensions are links, the area will be 
expressed in square links, and may be reduced to square 
chains by cutting o^ four places of decimals; ii Jive places 
be cut off, the remaining figures will be acres. If the deci- 
mal part of an acre be multiplied by 4, it will give the roods, 
and the resulting decimal, multiplied by 40, will give the 
perches. 

170. The difference of latitude, or the northing or soidhing 

II 




c 



114 Trigonometry. 

of a line, is the distance that one end is farther north or south 
than the other end. 

Thus, if ISTS be a meridian passing through the end A of 
the line AB, and EC be perpendicular to ]N"S, -^ 

then is AC the difference of latitude, or northing 
of AB. 

The departure, or the easting or westing of a 
line, is the distance that one end is farther east or 
west than the other end. 

Thus EC is tlie departure or westing of the line 
AB. 

It is evident that the distance, difference of lat- 
itude, and departure form a right-angled triangle, of which the 
distance is the hjpotheniise. 

The meridian distance of a point is the perpendicular let 
fall from the given point on some assumed meridian, and is 
east or west according as this point lies on the east or west 
side of the meridian. 

The meridian distance of a line is the distance of the middle 
point of that line from some assumed meridian. 

171. When a piece of ground is to be surveyed, we begin 
at one corner of the field, and go entirely around the field, 
measurino; the leno^th of each of the sides with a chain, and 
their bearings with a compass. 

Plotting a Survey. 
When a field has been surveyed, it is easy to draw a plan 
of it on paper. For this purpose, draw a line to represent the 
meridian passing through the first station ; then lay off an an- 
gle equal to the angle which the first side of the field makes 
with the meridian, and take the length of the side from a scale 
of equal parts. Through the extremity of this side draw a 
second meridian parallel to the first, and proceed in the same 
manner with the remaining sides. This method will be easily 
understood from an example. 

Example 1. 
Draw a plan of a field from the following courses and dis- 
tances, as given in the field-book. 





SUEVEYING. 


stations. 


Bearings. 


Distances. 


1 


N. 45° E. 


9.30 chains. 


2 


S. 60° E. 


11.85 " 


3 


S. 20° W. 


5.30 " 


4: 


S. 70° W. 


10.90 '^ 


5 


K 31° W. 


9.40 " 



115 




Draw NS to represent a meridian line; in NS take any con- 
venient point, as A, for the first station, and lay off an angle, 
NAB, equal to 45°, the bear- 
ing from A to B, which will 
give the direction from A to 
B. Then, from the scale of 
equal parts, make AB equal 
to 9.30, the length of the first a 
side; this will give the sta- 
tion B. Throuo'h B draw a 
second meridian parallel to 
NS ; lay off an angle of 60°, 
and make the line BC equal to 
11.85. Proceed in the same 
manner with the other sides. If the survey is correct, and the 
plotting accurately performed, the end of the last side, EA, 
will fall on A, the place of beginning. This plot is made on 
a scale of 10 chains to an inch. 

172. To avoid the inconvenience of drawing a meridian 
through each angle of the field, the sides may be laid down 
from the angles which they make with each other, instead of 
the angles which they make with the meridian. Reverse one 
of the bearings, if necessary, so that both bearings may run 
from the same angular point ; then the angle which any two 
contiguous sides make with each other may be determined 
from the following 

Rules. 

1. If both courses are north or south, and both east or west, 
subtract the less from the greater. 

2. If both are north or south, but one east and the other 
west, add them together. 



116 



TEIGONOMEtEY. 



3. If one is iiortli and the other south, but both east or west, 
subtract their sum from 180°. 

4. If one is north and the other south, one east and the other 
west, subtract their difference from 180°. 

Thus the angle CAB is equal to 
NAB-NAC. 

The angle CAD is equal to NAC 
4- NAD. 

The angle DAF is equal to 180° 
-(IsTAD + SAF). 

The angle CAF is equal to 180° 
-(SAF-NAC). 

In the preceding example we ac- 
cordingly find the angle 

ABC = 105°. DEA=101°. 

BCD = 100°. EAB = 104:°. 

CDE = 130°. 

With these angles the field may be plotted without drawing 
parallels. 




Example 2. 
The following field notes are given to protract the survey 



stations. 


Bearings. 


Distances. 


1 


K 50° 30^ E. 


16.50 chains. 


2 


S. 68° 15^ E. 


14.20 " 


3 


S. 9°45^E. 


8.45 " 


4 


S.21° O^W. 


6.84 '' 


5 


S. 73° 30' W. 


12.31 " 


6 


K 78° 15' W. 


9.76 " 


7 


K 15° 30' W. 


11.55 '' 



The Tkaveese Table. 
173. The accompanying traverse table shows the difference 
of latitude and the departure to two decimal places, for dis- 
tances from 1 to 100, and for bearings from 0° to 90°, at inter- 
vals of 15'. If the bearing is less than 45°, the angle will be 
found at the top of one of the pages of the table, and the dis- 
tance on the left and right of the page; the difference of lati- 



SUEVETING. 117 

tude will be found in the column headed Lot. at the top of the 
page, and the departure in the column headed Dejp. If the 
bearing is more than 45°, the angle will be found at the bot- 
tom of the page, and the difference of latitude will be found 
in the column marked Lat. at the bottom of the page, and the. 
departure in the other column. The latitudes and departures 
for different distances with the same bearing are proportional 
to the distances. Therefore the distances may be reckoned as 
tens, hundreds, or thousands, if the place of the decimal point 
in each departure and difference of latitude is changed ac- 
cordingly. 

Ex. 1. Kequired the latitude and departure for the distance 
43.29 and the bearing 27i°. 

On page 172, under 27^°, we find the following latitudes 
and departures, attention being paid to the position of the 
decimal points. 

Distance. ^W- ■^^^' Bep. 

43.00 38.23 19.69 

.29 .26 .13 

43.29 38.49 19.82 

Ex. 2. Eequired the latitude and departure for the distance 
5432 and the bearing 72^°. 

Distance. ^W- ^^^' Dep. 

5400 1624.00 5150.00 

32 9.62 30.52 

5432 1633.62 5180.52 

Ex. 3. Eequired the latitude and departure for the distance 
128.50 and the bearing 41i°. 

Ans. Latitude 95.87; Departure 85.56. 

The traverse table may be used not only for obtaining de- 
parture and difference of latitude, but for finding by inspection 
the sides and angles of any right-angled triangle ; for the lati- 
tude and departure form the two legs of a right-angled trian- 
gle, of which the distance is the hypothenuse, and the course 
is one of the acute angles. 

In this manner we find the latitude and departure for each 
side of the field given in Example 1, page 115, to be as in the 
following table : 



118 



Teigonometky. 



1 


Courses. 


Dis- 
tances. 


Latitude. | Departure. 


Cor. 
Lat. 


Cor. 
Dep. 


Balanced. | 


N. 


IS. 


E. 


w. 


N. 1 S. 


E. 


w. 


K45°E. 


9.30 


6.58 




6.58 






+ .01 


6.58 




6.59 




2 


S.60°E. 


11.85 




5.92 


10.26 




+ .01 


+ .01 




5.93 


10.27 




3 


S. 20° W. 


5.30 




4.98 




1.81 




-.01 




4.98 




1.80 


4 


S. 70°W. 


10.90 




3.73 




10.24 




-.01 




3.73 




10.23 


5 


N.3PW. 


9.40 


8.06 






4.84 




-.01 


8.06 






4.83 
16.86 


Perimeter 46.75 


14.64 


14.63116.84 


16.89 




14.64 


14.64 


16.86 



174. When a field lias been correctly surveyed, and the 
latitudes and departures accurately calculated, the sum of the 
northings should be equal to the sum of the southings, and the 
sum of the eastings equal to the sum of the westings. If the 
northings do not agree with the southings, and the eastings 
with the westings, there must be an error either in the survey 
or in the calculation. In the preceding example, the north- 
ings exceed the southings by one link, and the westings ex- 
ceed the eastings by five links. Small errors of this kind are 
unavoidable ; but when the error does not exceed one link to 
a distance of three or four chains, it is customary to distribute 
the error among the sides by the following proportion : 

As the perimeter of the field ^ 

Is to the length of one of the sides, 

/So is the error in latitude or departure, 

To the correction corresjponding to that side. 

This correction, when applied to a column in which the sum 
of the numbers is too small, is to be added j but if the sum of 
the numbers is too great, it is to be subtracted. 

We thus obtain the corrections in columns 8 and 9 of the 
preceding table ; and applying these corrections, we obtain the 
balanced latitudes and departures, in which the sums of the 
northings and southings are equal, and also those of the east- 
ings and westings. 

As the computations are generally carried to but two deci- 
mal places, the corrections of the latitudes and departures are 
only required to the nearest link, and these corrections may 
often be found by mere inspection without stating a formal 
proportion. Thus, in the preceding example, since the depart- 
ures require a correction of five links, and the field has five 
sides which are not very unequal, it is obvious that we must 
make a correction of one link on each side. 



SuKVEYINa. 119 

It is the opinion of some surveyors that when the error in 
latitude or departure exceeds one link for every five chains of 
the perimeter, the field should he resurveyed ; but most sur- 
veyors do not attain to this degree of accuracy. The error, 
however, should never exceed one link to a distance of two or 
three chains. 

175. To find the area of the field. 
Let ABODE be the field .^.. 

to be measured. Through i -r 

A, the most western station, [ 

draw the meridian NS, and 

upon it let fall the perpen- i / ^"". 

diculars BF, CG, DH, EL g-LZ-- \c 

Then the area of the re- j\ / 

quired field is equal to j \ / 

FBCDEI-(ABFh-AEI). hI--\- -- ..^jy 

But FBCDEI is equal to i \ ,.,.-^-''" 

the sura of the three trape- -j-l a,--"' 

zoids FBCG, GCDH, HDEL | ^ 

Also, if the sum of the "^ 
parallel sides FB, GO be multiplied by FG, it will give twice 
the area of FBCG (Art. 104). The sum of the sides GO, DH, 
multiplied by GH, gives twice the area of GCDH ; and the 
sum of HD, IE, multiplied by HI, gives twice the area of 
HDEL 

Now BF is the departure of the first side, GC is the sum of 
the departures of the first and second sides, HD is the alge- 
braic sum of the three preceding departures, IE is the algebraic 
sum of the four preceding departures. Then the sum of the 
parallel sides of the trapezoids is obtained by adding together 
the preceding meridian distances two by two ; and if these 
sums are multiplied by FG, GH, &c., which are the corre- 
sponding latitudes, it will give the double areas of the trape- 
zoids. 

176. It is most convenient to reduce all these operations 
to a tabular form, according to the following 

Rule. 
Having arranged the balanced latitudes and dejpartures in 



120 



Teigonometet. 



their aj)proj>riate columns, draw a meridian through the most 
eastern or western station of the survey, and, calling this the 
first station, form a column of double meridian distances. 

The double meridian distance of the first side is equal to its 
departure / and the double meridian distance of any side is 
equal to the double meridian distance of the preceding side, 
plies its departure, plus the departure of the side itself. 

Multiply each double meridian distance by its correspond- 
ing northing or southing, and place the product in the column 
of north or south areas. The difference between the sum of the 
north areas and the sum of the south areas will be double the 
area of the field. 

It must be borne in mind that by the iQvm plus in this rule 
is to be understood the algebraic sum. Hence, when the 
double meridian distance and the departure are both east or 
both west, they must be added together; but if one be east 
and the other west, the one must be subtracted from the 
other. 

The double meridian distance of the last side should always 
be equal to the departure for that side. This coincidence af- 
fords a check against any mistake in forming the column of 
double meridian distances. 

The preceding example will then be completed as follows: 





N. 


s. 


E. 


w. 


D.M.D. 


N. Areas. 


S. Areas. 


1 


6.58 




6.59 




6.59 


43.3622 




2 




5.93 


10.27 




23.45 




139.0585 


3 




4.98 




1.80 


31.92 




158.9616 


tt 




3.73 




10.23 


19.89 




74.1897 


5 


8.06 






4.83 


4.83 


38.9298 






82.2920 


372.2098 



Twice the figure FBCDEI is 372.2098 square chains. 
Twice the figure FBAEI is 82.2920 " 

The difference is 289.9178 " 

Therefore the area of the field is 144.9589 square chains, or 
14.49589 acres, which is equal to 14 acres, 1 rood, 39 perches. 

Ex. 2. It is required to find the contents of a tract of land 
of which the following are the field notes : 



Surveying. 



121 



Sta- 
tions. 


Bearings. 


Distances. 


1 


N. 50° 30' E. 


16.50 chains. 


2 


S. 68° 15' E. 


14.20 " 


3 


S. 9° 45' E. 


8.45 " 


4 


S. 21° O'W. 


6.84 " 


5 


S. 73° 30' W. 


12.31 " 


6 


K 78° 15' W. 


9.76 '^ 


7 


]NM7° O'W. 


11.64 " 




Calculation. 



1 


Courses. 


Dif. Lat. 1 Departure. 


c. 


Balanced. 


D. M. 


N. Areas. 


S. Areas. 


T 


Diot. 1 ^_ 


s. 


E. 


W. 


N. 1 S. 1 E. 1 \V. 




N. 50° 30' E. 


16.50 10.50 




12.73 




.03 


10.47 




12.70 




12.70 


132,9690 




2 


S. 6SO 15' E. 


14.20: 


5.26 


13.19 




.03 




5.29 


13.16 




38.56 




203.9824 


3 


S. 9° 45' E. 


8.45 


8.33 


1.43 




.01 




8.34 


1.42 




53.14 




443.1876 


+ 


S. 21° O'W. 


6.84 


6.39 




2.45 


.01 




6.40 




2.46 


52.10 




333.4400 


5 


S. 73° 30' W. 12.31 1 


3.50 




11.80 


.02 




3.52 




11.82 


37.82 




133.1264 


fi 


N. 78° 15MV. 1 9.76! 1.99 






9.56 


.02 


1.97 






9.58116.42 


32.3474 




7 


N. 17° O'W. 11.64 11.13 






3.40 


.02 


11.11 






3.42 3.42 


37.9962 




|79. 70 23.62 


23.48 


27.35 


27.21 




23.55|23.55 


27.28 


27.28| 


203.3126 


1113.7364 


1 Error ,14 


Error .14 




203.3126 



Ans.. 45 A., 2 E., 3 P. 



2)910.4238 
455.2119. 



Ex. 3. Required tlie area of a 
tract of land of which tlie follow- 



ing 


are the field notes : 




sta- 
tions. 


Bearings, 


Distances. 


1 


K 58° 45' E. 


19.84 cha 


ms. 


2 


N. 39° 30' E. 


10.45 ' 




3 


S. 45° 15' E. 


37.26 ' 




4 


S. 52° 30' W. 


21.53 ' 




5 


S. 34° O'E. 


9.12 ' 




6 


S. 66° 15' W. 


27.69 ' 




7 


N. 12° 45' E. 


24.31 ' 




8 


N. 48° 15' W. 


24.60 ' 






Ans., 130 A., 2 R., 23 P. 

Ex. 4. Required the area of a piece of land from the follow- 



ing field notes 



122 



Teigonometey. 



Stations. 


Bearings. 


])istances. 


1 


N. 5° 15' E. 


15.17 c 


iiaiiis. 


2 


N. 45° 45' E. 


16.83 




3 


K32° O'W. 


14.26 




4 


IS^. 88° 30' E. 


19.54 




5 


S. 28° 15' E. 


17.92 




6 


S. 40° 45' W. 


9.71 




7 


S. 31° 30' E. 


22.65 




8 


S.14° O'W. 


18.39 




9 


S. 82° 45' W. 


24.80 




10 


N. 23° 15' W. 


26.31 





Ans., lid A., 1 E., 16 P. 
Ex. 5. Eequired the area of a field from the following notes 



Station.?. 


Bearings. 


Distances. 


1 


K 32° 15' E. 


28.74 c] 


lains. 


2 


K 17° 45' E. 


21.59 




3 


S. 81° 30' E. 


13.38 




4 


S. 9°45'W. 


11.92 




5 


S.43° O'E. 


19.65 




6 


K 25° 30' E. 


17.26 




7 


S. 78° 15' E. 


18.87 




8 


S. 5°45'W. 


31.41 




9 


S. 37° 30' W. 


26.13 




10 


N. 69° O'W. 


23.86 




11 


S. 74° 15' W. 


20.91 




12 


N. 27° 30' W. 


23.20 





Ans., 304 A., 2 K., 9 P. 
Ex. 6. Peqnired the area of a field from the following notes 



Station.s. 


Bearing.s. 


Distances. 


1 


N. 36° 15' E. 


24.73 chains. 


2 


K 7°45'E. 


11.58 " 


3 


K 79° 30' E. 


15.39 " 


4 


S. 86° 45' E. 


20.56 " 


5 


S. 12° 15' W, 


18.14 '' 


6 


S.25° O'E. 


21.92 " 


7 


S. 58° 30' W. 


29.27 " 


8 


K 34° O'W. 


19.81 " 


9 


K 81° 15' ^Y. 


21.24 " 



^72^., 179 A., IE., 6 r. 



Sfeyetii^g. 



123 



177. The field notes from ^vliich the area is to be com- 
puted may be imperfect. There may be obstacles which pre- 
vent the measuring of one side, or the notes may be defaced 
so as to render some of the numbers illeo^ible. If the bearino:s 
and lengths of all the sides of a field excejot one are given, the 
remaining side may easily be found by calculation. For the 
difference between the sum of the northings and the sum of 
the southiugs of the given sides will be the northing or south- 
ing of the remainiug side ; and the difference between the sum 
of the eastino:s and the sum of the westins^s of the eiven sides 
will be the easting or westing of the remaining side. Having, 
then, the difference of latitude and departure of the required 
side, its length and direction are easily found by Trigonome- 
try (Art. 54). 

Ex. Given the bearings and lengths of the sides of a tract 
of land as follows : 



Stations. 


Bearings. 


Distances. 


1 


K 18° 15^ E. 


8.93 chains. 


2 


K 79° 45^ E. 


15.64 " 


3 


S. 25° O'E. 


14.27 " 


4 


Unl'nown. 


TJnhwwn. 


5 


1^. 87° 30' W. 


18.52 chains. 


6 


K. 41° 15' W. 


12.18 " 



Required the bearing and distance of the fourtli side. 

Ans., S. 15° 33' E., distance 8.62 chains. 

178. There is another method of finding the area of a field 
which may be practised when great accuracy is not required. 
It consists in first drawing a plan of the field, as in Art. 171, 
then dividing the field into 
triangles by diagonal lines, 
and measuring the bases and 
perpendiculars of the triangles 
upon the same scale of equal 
parts by which the plot was 
drawn. Thus, if we take Ex. 
1, and draw the diagonals AC, 
AD, the field will be divided 
into three triangles, whose area 
is easily found when we know 




124: 



Tkigonometkt. 



the diagonals AC, AD, and the perpendiculars BF, DG, EH. 
The diagonal AC is found by measurement upon the scale of 
equal parts to be 16.87 ; the diagonal AD is 15.67 ; the perpen- 
dicular BF is 6.30 ; DG is 4.92; and EH is 6.42. Hence 

the triangle ABC = 16.87x3.15= 53.14 
" " ADC = 16.87x2.46= 41.50 
" " ADE = 15.67x3.21= 50.30 

the figure ABCDE =144.94 sq. chains. 

This method of finding the area of a field is very expedi- 
tious, and when the plot is carefully drawn, may afford results 
sufficiently precise for many purposes. 

179. To survey an irregular houndary by means of off- 
sets. 

When the boundaries of a field are very irregular, like a 
river or lake shore, it is generally best to run a straight line, 
coming as near as is convenient to the true boundary, and 
measure the perpendicular distances of the prominent points 
of the boundary from this line. 

Let ABCD be a piece of land to be surveyed ; the land be- 
ing bounded on the east by a lake, and on the west by a creek. 
We select stations A, B, C, 
D, so as to form a polygon 
which shall embrace most of 
the proposed field, and find 
its area. We then measure 
perpendiculars aa^, l)b\ cg\ 
&c., as also the distances Aa, 
a5, 1)0^ &c. Then, consider- 
ing the spaces Kaa' ^ aWa'^ 
&c., as triangles or trapezoids, 
their area may be computed ; 
and, adding these areas to the 
figure ABCD, we shall obtain 
the area of the proposed field nearly. 

180. To determine the 'hearing and distance from one point 
to another hy means of a series of triangles. 

When it is required to find the distance between two points 
remote from each other, we form a series of triangles such that 




Surveying. 



125 



the first and second triangles may Lave one side in common ; 
the second and third, also, one side in common ; the third and 
fourth, &G. We then measure one side of the first triangle 
for a base line, and all the angles in each of the triangles. 
These data are sufficient to determine the length of the sides 
of each triangle ; for in the first triangle we have one side and 
the angles to find the other sides. When these are found, we 
shall have one side and all the angles of a second triangle to 
find the other sides. In the same manner we may calculate 
the dimensions of the third triangle, the fourth, and so on. 
We shall illustrate this method by an example taken from the 
Coast Survey of the United States. 

The object here is to make a survey of Chesapeake Bay and 
its vicinity; to determine with the 
utmost precision the position of the 
most prominent points of the country, 
to which subordinate points may be 
referred, and thus a perfect map of 
the country be obtained. According- 
ly, a level spot of ground was select- 
ed on the eastern side of the bay, on 
Kent Island, where a base line, AB, 
of more than five miles in length, 
was measured with everj^ precaution. 
A station, C, was also selected upon 
the other side of the bay, near An- 
napolis, so situated that it was visible 
from A and B. The three angles of 
the triangle ABC were then meas- 
ured with a large theodolite, after 
which the length of BC may be com- 
puted. A fourth station, D, is now 
taken on the western shore of the bay, visible from C and B, 
and all the angles of the triangle BCD are measured, when the 
line BD can be computed. A fifth station, E, is now taken on 
an island near the eastern shore, visible both from B and D, and 
all the angles of the triangle BDE are measured, when DE can 
be computed. Also, all the angles of the triangle DEE are 
measured, and EF is computed. Then all the angles of the tri- 




126 Teigonometrt. 

angle EFG are measured, and FG is computed. So, also, all 
tlie angles of the triangle FGH are measured, and GH is com- 
puted; and thus a chain of triangles maj be extended along the 
entire coast of the United States. To test the accuracy of the 
work, it is common to measure a side in one of the triangles 
remote from the first base, and compare its measured length 
with that deduced by computation from the entire series of tri- 
angles. This line is called a hase of verification. Such a base 
has been measured on Long Island ; and, indeed, several bases, 
have been measured on different points of the coast. These 
are all connected bj a triangulation, and thus the length of a 
side in any triangle may be deduced from more than one base 
line, and the agreement of these results is a test of the accuracy 
of the entire work. Tlius the length of one of the sides of a 
triangle which was twelve miles, as deduced from the Kent 
Island base, differed only twenty inches from that derived 
from the Long Island base, distant two hundred miles. 

The superiority of this method of surveying arises from the 
circumstance that it is necessary to measure but a small num- 
ber of base lines along a coast of a thousand or more miles in 
extent ; and for these the most favorable ground may be se- 
lected anywhere in the vicinity of the system of triangles. 
All the other quantities measured are angles ; and the pre- 
cision of these measurements is not at all impaired by the in- 
equalities of the surface of the ground. Indeed, mountainous 
countries afford peculiar. facilities for a trigonometrical survey, 
since they present heights of ground visible to a great distance, 
and thus permit the formation of triangles of very large di- 
mensions. 

181. To divide an irregular j^iece of land into any two 
given parts. 

We first run a line, by estimation, as near as may be to the 
required division line, and compute the area thus cut off. If 
this is found too large or too small, we add or subtract a tri- 
angle, or some other figure, as the case may require. Sup- 
pose it is required to divide the field ABCDEFGHI into two 
equal parts, by a line IL, running from the corner I to the 
opposite side CD. We first draw a line from I to D, and com- 
pute the area of the part DEFGHI; and, knowing the area 



SlJEVEYING. 



127 




of the entire field, we learn the area which must be contained 
in the triangle DIL, in order that 
IL may divide the field into two 
equal parts. Having the bearings 
and distances of the sides DE, EF, 
&c., we can compute the bearing 
and distance of DI. Thus the an- fi^i 
gle IDK is known ; and, having 
the hjpothennse ID, we can com- 
pute the length of the perpendicu- 
lar IK let fall on CD. Now the 
base of a triangle must be equal 
to its area divided by half the al- 
titude. Hence, if we divide the G- E 
area of the triangle DIL by half of IK, it will give DL. 

In a similar manner we might proceed if it was required to 
divide a tract of land into any two given parts. 

Variation of the Needle. 

182. The line indicated by a magnetic needle, when free- 
ly supported and allowed to come to a state of rest, is called 
the magnetic meridian. This does not generally coincide with 
the astronomical meridian^ which is a true north and south 
line. 

The angle which the magnetic meridian makes with the 
true meridian is called the variation of the needle.^ and is said 
to be east or west, according as the north end of the needle 
points east or west of the north pole of the earth. 

The variation of the needle is diiferent in different parts of 
the earth. In some parts of the United States it is 10° west, 
and in others 10° east, while at other places the variation has 
every intermediate value. Even at the same place, the varia- 
tion does not remain constant for any length of time. Hence 
it is necessary frequently to determine the amount of the vari- 
ation, which is easily done when we know the position of the 
true meridian. The latter can only be determined by astro- 
nomical observations. The best method is by observations of 
the pole star. If this star were exactly at the pole, it would 
always be on the meridian ; but, being at a distance of nearly 



128 



Teigonometrt. 



a degree and a half from the pole, it revolves about the pole 
in a small circle in a little less than twentj-four hours. In 
about six hours from its passing the meridian above the pole, 
it attains its greatest distance west of the meridian ; in about 
six hours more it is on the meridian beneath the pole ; and in 
about six hours more it attains its greatest distance' east of the 
meridian. If the star can be observed at the instant when it 
is on the meridian, either above or below the pole, a true north 
and south line may be obtained. 

183. The following table shows the time of the pole star's 
passing the meridian above the pole for every fifth day of the 
year : 



1st Day. 



6th Dav. 



January 

February. . . 

ilarch 

April 

May 

June 

July 

August 

September.. 

October 

November . . 
December. . . 



6 30 P.M. 
4 28 " 
2 37 " 
31 " 

10 33 A.M. 
8 32 " 
6 3i " 
4 33 " 
2 32 ' = 
34 " 

10 32 P.M. 
8 34 '^ 



b. m. 
6 10 P. 51. 
4 8" 
2 18 " 
12 " 

10 14 A.M. 
8 12 '• 
6 15 " 
4 13 " 
2 12 " 
14 " 

10 12 P.M. 
8 14 " 



5 51 
3 48 
1 58 

11 52 
9 54 
7 53 
5 55 
3 54 
1 52 

11 54 
9 53 
7 54 



P.M. 
A.M. 



P.:iL 



31P.M. 
29 " 
33 " 
32 
35 
33 



.A.M. 



35 P. M. 
33 >• 
35 " 



h. m. 
5 11 
3 9 
1 19 

11 13 
9 15 
7 14 
5 16 
3 15 
1 13 

11 15 
9 13 
7 15 



P.M. 



P.M. 



51 P.M. 

49 " 
59 " 

53 A.M. 

55 '• 

54 " 

56 " 

55 " 
53 " 
55 P.M. 
53 " 
55 " 



If the pole star passes the meridian in the daytime, it can- 
not be observed without a good telescope ; but 11^- SS""* after 
the dates in the above table, the star will be on the meridian 
deloio the pole, and during the whole year, except in summer, 
the pole star may be seen with the naked eye on the merid- 
ian either above or below the pole. These observations are 
best made with a theodolite, but they may be made with a 
common compass. At 5^- 59°^ after the dates in the above 
table, the star will have attained its greatest distance west of 
the meridian ; and 5^' 69"^' before these dates, it will be at its 
greatest distance east of the meridian. In summer, therefore, 
we may observe the greatest eastern elongation of the pole 
star, at which time the star is 1° 43' east of the true meridian 
for all places in the neighborhood of !N"ew York. Making this 
allowance, a true meridian is easily obtained ; after which, the 
variation of the needle is determined by placing a compass 
upon this line, turning the sights in the same direction, and 
noting the angle shown by the needle. 

The following table shows the angle which the plane of the 



Surveying. 



129 



meridian makes with a vertical plane passing throngli the pole 
star, when at its greatest eastern or western elongation, for 
any latitude from 30° to 44°. 



Lat. 30° 


Lat. 32° 


Lat. 34° 


Lat. 36° 


Lat. 38° 


Lat. 40° 


Lat. 42° 


1° 30' 


1° 32' 


1° 34' 


i°3r 


1°39' 


1°42' 


1°45' 



Lat. 44° 
1° 49' 



184. The variation of the needle, in 1885, for several parts 
of the United States, was as follows : 



Bangor, Me 16°. 

Portland, Me 13°. 

Boston, Mass 11°. 

Burlington, Yt 12°. 

Kew Haven, Conn 9°. 

New York City 8°. 

Albany, N. Y 10°. 

Rochester, K Y 6°. 



Buffalo, X. Y 

Pliiladelpliia, Pa. . . 

Pittsburgh, Pa 

Erie, Pa 

Baltimore, Md 4°. 

Washington, D. C 3°. 

Richmond, Ya 2°. 

Norfolk, Ya 3°. 

Lynchburg, Ya 1°. 

Raleigh, N. C 1°. 

TVilmington, X. C 0°. 

Charleston, S. C 0°. 

Columbia, S. C 1°. 

Savannah, Ga 1°. 

Augusta, Ga 2°. 

Athens, Ga 2°. 

Macon, Ga 2°. 



67 W. 

25 AY. 
78 W. 
41 W. 
19 W. 
03 TT. 
19 TT. 

38 W. 
73 W. 
70 W. 
72 W. 

39 W. 
47 W. 
87 W. 
90 W. 

26 W. 
23 W. 
13 W. 
61 W. 
17 E. 



2°. 
5°. 
Tuscaloosa, Ala 5°. 



St. Augustine, Fla. 
Pensacola, Fla. . . . 



Montgomery, xVla 4°. 

Yicksburg, Miss 6°. 

Natchez, Miss 6°. 

New Orleans, La 6°. 

Galveston, Texas 7°.83 E. 



Austin, Texas 8°. 

Little Rock, Ark 7°. 

Knoxville, Tenn 1°. 

Nashville, Tenn 4°. 

Louisville, Ky 3°. 

Wheeling, W. Ya 0°. 

Cleveland, O 1°. 

Columbus, 0°. 

Cincinnati, 1°. 

Indianapolis, lud 2°. 

Terre Haute, Ind 3°. 

Chicago, 111 4°. 

Springfield, 111 5°. 

Cairo, 111 5°. 

St. Louis, Mo 

Jefferson City, Mo. . . 

Kansas City, Mo 

Detroit, Mich 

Kalamazoo, Mich.. . . 
Grand Haven, Mich.. 

Milwaukee, AYis 

Madison, Wis 6°. 

Dubuque, la 6°. 

Iowa City, la 7°. 

St. Paul, Minn 9°. 

Minneapolis, Minn 9°. 

Duluth, Minn 10°. 

Denver, Col 14°. 

Salt Lake City, Utah. . . 16°. 

San Francisco, Cal 16°. 

San Diego, Cal 13°. 

Portland, Oregon 22°. 

Olympia, Wash. Ter. . . 21°. 



65 
72 
86 
01 

22 W. 
51 W. 
64 E. 
95 E. 
48 E. 

85 E. 
04 E. 
48 E. 
62 E. 
00 E. 

23 E. 

86 E. 
35 W. 
39 E. 
15 E. 
12 E. 
15 E. 
47 E. 
71 E. 
94 E. 



E. 

E. 

E. 

E. 

E. 

E. 
33 E. 
61 E. 



185. The following table shows the annual change in the 
variation of the needle. Where the variation is Westerly it 
is increasing; where it is Easterly it is decreasinir. 

I 



130 



Tkigonometey. 



Portland, Me 1'.2 

Burlington, Vt 5'. 8 

Boston, Mass 2'. 5 

New Haven, Conn 4'.1 

New York City 2'.6 

Albany, N. Y 3'. 4 

Buffalo, KY 4'.8 

Philadelpliia, Pa 5'. 3 

Washington, D. C 3'.0 

Charleston, S. C 2'. 7 



Savannah, Ga 3'. 3 

New Orleans, La 3'. 7 

Cleveland, 2'.2 

Cincinnati, 3'.8 

Detroit, Mich 2'.8 

Chicago, 111 4'.6 

St. Louis, Mo 3'.0 

Madison, Wis 3'. 9 

St. Paul, Minn 3'.0 

Denver, Col 1'.6 



LEVELLING., 

186. Levelling is the art of determiniDg the difference of 
level between two or more places. 

The surface of an expanse of tranquil water, or any surface 
parallel to it, is called a level surface. Points situated in a 
level surface are said to be on the same level, and a line traced 
on such a surface is called a line of true level. 

On account of the globular figure of the earth, a level sur- 
face is not a plane surface. It is nearly spherical; and in the 
common operations of levelling it is regarded as perfectly so. 
Hence every point of a level surface is regarded as at the same 
distance from the centre of the earth ; and the difference of 
level of two places is the difference between their distances 
from the centre. 

A line of apjparent level is a straight line tangent to the sur- 
face of the earth. 

Thus, if AB represent the surface of the ocean, the two 
places A and B are said to be on the same level ; but if AD 
be drawn tangent to the arc AB at A, then AD is a line of 
apparent level. _^ 

This is the line which is indicated by a 
levelling instrument placed at A. The theod- 
olite may be employed for tracing horizontal 
lines; but if nothing further were required, 
there would be no occasion for graduated cir- 
cles, and several parts of the theodolite might 
be dispensed with. A levelling instrument, therefore, usually 
consists of a large spirit level attached to a telescope, mounted 
upon a stand in a manner similar to the theodolite. 




Surveying. 



131 




187. The surveyor should also be provided with a pair of 
levelling staves. A levelling staff consists of a 
rectangular bar of wood six feet in length, di- 
vided to inches and sometimes tenths of an 
inch, and having a groove running its entire 
length. A smaller staff of the same length, 
called a slide, also divided into inches, is in- 
serted in this groove, and moves freely along it. 

At the upper end of the slide is a rectangu- 
lar board called a vane, AB, about six inches 
wide. The vane is divided into four equal parts by two lines, 
one horizontal and the other vertical. Two opposite parts of 
the vane are painted white, and the other two black, in order 
that they may be distinguished at a great distance. 

To find the difference of level letween any two points. 

188. Set up the levelling staves perpendicular to the hori- 
zon, and at equal distances from the levelling instrument. 
Having adjusted the level by means of the proper screws, turn 
the telescope to one of the staves, and direct an assistant to 
slide up the vane until the line AB coincides with the centre 
of the telescope, and note the height of this line from the 
ground. Turn the telescope to the other staff, and repeat the 
same operation. Level in the same manner from the second 
station to the third, from the third to the fourth, &c. Then 
the difference between the sum of the heights at the back sta- 
tions and at the forward stations will be equal to the difference 
of level between the first station and the last. 

If we wish to level from A to E, we set up the staves at a 
convenient distance, 
AC, and midway be- 
tween them place the 
level B. Observe 
where the line of lev- 
el, FGr, cuts the rods, and note the heights AF, CG. Their 
difference is the difference of level between the first and sec- 
ond stations. Take up the level and place it^at D, midway 
between the rods C and E, and observe where the line of level, 
HI, cuts the rods, and note the heights CH, EI. 




132 



Trigonometry. 



Then FA— CG = the ascent from A to C, 

and CH — EI =the ascent from C to E. 

Therefore (FA + CIl)— (CG + EI)=:the entire ascent from 
A to E ; and in the same manner we maj find the difference 
of level for any distance ; that is, the difference between the 
sum of the heights at the hack stations and at the forward sta- 
tions is equal to the difference of level letween the first station 
and the last. 

189. The following is a copy of the field notes for running 
a level from A to E : 

Back sights. Fore sights. 



Feet. 


Inches. 





4 


5 


10 


4 


2 


5 


6 


4 


11 


4 


7 


6 


1 



Feet. 

3 

5 
4 
1 
3 
1 
2 



Inches. 

2 
7 
3 
2 
2 
3 




Sum 31 5 Sum 20 7 

The back sights being greater in amount than the forward 
sights, it is evident that E is higher than A by 10 feet 10 
inches. 

The heights indicated by the levelling staves are sometimes 
read off by the assistant, but it is better for the observer to 
read off the quantities himself througli the telescope of his 
levelling instrument. This may easily be done provided the 
graduation of the staff is perfectly distinct ; and in that case it 
is only necessary to rely upon the assistant to hold the staff 
perpendicularly. To enable him to do this, a small plummet 
is suspended in a groove cut in the side of the staff. 

190. It must be observed tlmt the lines GF, HI are lines 
of ajpparent level, and 
not of true level ; nev- 
ertheless, we shall ob- 
tain the true differ- 
ence of level between 
A and E by this method if the levelling instrument is placed 
midway between the levelling staves, because the points G and 




SUEYEYING. 



133 



F will in that case be at equal distances from the earth's cen- 
tre. If the level is not placed midway between the staves, 
then we must apply a correction for the difference between 
the true and apparent level. 

191. To find the difference leticeen the true and apjparent 
level. 

Let C be the centre of the earth, AB a portion of its surface, 
and AD a tano-ent to the earth's surface at 
A; then BD is the difference between the 
true and apparent level for the distance AD. 

ISTow, by Geom., Prop. 11, B. lY., 

CD^rrACHADl 

Hence 
and 



CD =VaC^ + AD^ 




BD =VACHAD-^-BC. 
If we put R=:BC, the radius of the earth, 
^=AD, 
and A = BD, the difference between the true and ap- 

parent level, we shall have 

that is, to find the difference between the true and apparent 
level for any distance, add the square of the distance to the 
square of the eartKs radius^ extract the square root of the sum, 
and sid)tract the radius of the earth. 

If BD represent a mountain, or other elevated object, then 
AD will represent the distance at which it can be seen in con- 
sequence of the curvature of the earth. 

Ex. 1. If the diameter of the earth be 7912 miles, and if 
Mount JEtna can be seen at sea 126 miles, what is its heio>ht? 

Ans., 2 miles. 

Ex. 2. If a straight line from the summit of Chimborazo 
touch the surface of the ocean at the distance of 179 miles, 
what is the height of the mountain ? Ans., 4.05 miles. 

From the preceding formula we obtain 

==RH2RA + A^, 
that is, d' = ^Wi + h\ 

But in the common operations of levelling, h is very small in 
comparison with the radius of the earth, and h' is very small 



134 Teigonometey. 

in comparison with 2RA. If we neglect the term U^ we have 

whence A — -p; 

that is, the difference hetween the true and apparent level is 
nearly equal to the square of the distance divided hy the diam- 
eter of the earth. 

Ex. 1. What is the difference between the true and apparent 
level for one mile, supposing the diameter of the earth to be 
7912 miles ? A^is., 8.008 inches, or 8 inches nearly. 

Ex. 2. What is the difference between the true and apparent 
level for half a mile ? Ans., 2 inches. 

d" . . 

In the equation h=^y^, since 2E. is a constant quantity, A 

varies as d"^ ; that is, the difference hetv^een the true and appar- 
ent level varies as the square of the distance. 

Hence, the difference for 1 mile being 8 inches. Ft. ins. 
the difference for 2 miles is 8 x 2'= 32 inches = 2 8 

3 '' 8x3'= 72 '' =60 

4 " 8x4^ = 128 " =10 8 

5 " 8x5^ = 200 " =16 8 

6 " 8x6^ = 288 " =21 0, &c. 

Topographical Maps. 

102. It is sometimes required to determine and represent 
upon a map the undulations and inequalities in the surface of 
a tract of land. Such a map should give a complete view of 
the ground, so as to afford the means for an appropriate loca- 
tion of buildings or extensive works. For this purpose, we 
suppose the surface of the ground to be intersected by a num- 
ber of horizontal planes, at equal distances from each other. 
The lines in which these planes meet the surface of the ground, 
being transferred to paper, will indicate the variations in the 
inclination of the ground ; for it is obvious that the curves will 
be nearer together or farther apart, according as the ascent is 
steep or gentle. 

Thus, let ABCD be a tract of broken ground, divided by a 
stream, EF, the ascent being rapid on each bank, the ground 



(( 


a 


<( 


a 


(£ 


u 


U 


u 



SUEVETING. 



135 



swelling to a liill at G, and also at H. It is required to repre- 
sent these inequal- 



A. 



B 






ii^i 












»^ 







<i^\ 



WMMy 



ities upon paper, 
so as to give an 
exact idea of the 
face of the groun d . 
The lowest point 
of the ground is 
at F. Suppose the 
tract to be inter- 
sected by a hori- 
zontal plane four 
feet above F, and 
let this plane in- 
tersect the surface '^ E D 
of the ground in the undulating lines marked 4, one on each 
side of the stream. Suppose a second horizontal plane to be 
drawn eight feet above F, and let it intersect the surface of the 
ground in the lines marked 8. Let other horizontal planes be 
drawn at a distance of 12, 16, 20, 21, &c., feet above the point 

F. The projection of these lines of level upon paper shows 
at a glance the outline of the tract. We perceive that on the 
right bank of the stream the ground rises more rapidly on the 
upper than on the lower portion of the map, as is shown by 
the lines of level being nearer to one another. On the right 
bank of the stream the ascent is uninterrupted until we reach 

G, which is the summit of the hill. Beyond G the ground 
descends again toward B. On the left bank of the stream the 
ground rises to H ; but toward A the level line of 12 feet di- 
vides into t\vo branches, and between them the ground is 
nearly level. 

193. The surveys requisite for the construction of such a 
map may be made with a theodolite or common level. 

The object is to trace a series of level lines upon the surface 
of the ground. For this purpose we may select any point on 
the surface of a hill, place the level there, and run a level line 
around the hill, measuring the distances, and also the angles, 
at every change of direction. "We may then select a second 
point at any convenient distance above or below the former, 



136 



Teigonometey. 



and trace a second level line around the hill, and so on for as 
many curves as may be thonglit necessary. Such a method, 
liowever, would not always be most convenient in practice. 

194. The following method may sometimes be preferable: 
Set up the level on the summit of tlie hill at G, and fix the vane 
on the levelling staff at an elevation of four feet in addition to 
the height of the telescope above the ground. Then direct an 
assistant to carry the levelling staff (holding it in a vertical 
position) toward K, till he arrives at a point, as a^ where the 
vane appears to coincide with the cross wires of the telescope. 
This will determine one point of the cmwe line four feet be- 
low G. The assistant may then proceed to the line GB, and 
afterward to GL, moving backward or forward in each of those 
directions till he finds points, as d and ^, at which the vane 
coincides with the cross wires of the telescope. The horizontal 
distance between G and a, G and d^ G and g^ must then be 
measured. 

If the levelling staff is sufficiently long, the vane may be 
fixed on it at the height of eight feet, in addition to the height 
of the telescope at G; and the assistant, placing himself in the 
directions GK, GB, GL, must move till the vane appears to 
coincide with the cross wires as before. The horizontal dis- 
tances ab, de, gh^ must then be measured, and stakes driven 
into the ground at h, e, and A. 

The level must now be removed to Z>; and the vane being 
fixed on the staff 
at a heigh 



t equal 
to four feet, to- 
gether with the 



height of the in- 
strument from 
the ground at &, 
the assistant must 
proceed in the di- 
rection ^K, and 
stop at c wdien 
the vane coincides 
with the cross 
wires: then the 




SUKYETING. 137 

horizontal distance of c from h must be measm-ed. In a like 
manner, the operations may be continued from h or c as far as 
necessary toward K; then, commencing at e^ and afterward at 
A, they may be continued in the same way toward B and L 
respectively. The angles which the directions GK, GB, GL 
make with the magnetic meridian being found with the com- 
pass, these directions may be represented on paper. Then the 
measured distances G(2, ab^ &c. ; Gc?, de^ &c. ; G^, gli^ &c., be- 
ing set off on those lines of direction, curves drawn tlirough cc, 
d^g\ &, ^, A ; c^f^ 7j, &c., will show the contour of the hill. 

The map is shaded so as to indicate the hills and slopes by 
drawing fine lines, as in the figure, perpendicular to the hori- 
zontal cm-ves. 

195. Another method, which may often be more conven- 
ient than either of the preceding, is as follows : From the sum- 
mit of the hill measure any line, as GK, and at convenient 
points of this line let stakes be driven, and their distances from 
G be carefully measured. Then determine the difference of 
level of all these points ; and if the assumed points do not fall 
upon the horizontal curves which are required to be delineated, 
we may, by supposing the slope to be uniform from one stake 
to another, compute by a proportion the points where the hori- 
zontal curves for intervals of four feet intersect the line GK. 
The same may be done for the lines GB and GL, and for other 
lines, if they should be thought necessar3\ 

196. If the surface of the ground is gently undulating, it 
may be more convenient to run across the tract a number of 
lines parallel to one another. Drive stakes at each extremity 
of these lines, and also at all the points along them where 
there is any material change in the inclination of the ground, 
and find the difference of level between all these stakes, and 
their distances from each other. Then, if we wish to draw 
upon a map the level lines at intervals of 4, 6, or 10 feet, we 
may compute in the manner already explained the points. 
where the horizontal curves intersect each of the parallel lines. 
The curve lines are then to be drawn through these points, ac- 
cording to the judgment of the surveyor. 

197. If it is required to draw 2i 2^^"ofile of the ground, for 
example, from G to K, di-aw a straight line, GTv, to represent 



138 



Tkigonometet. 




a horizontal line to which the heights are referred, and set off 

GV, Q'h', G'g', &c., equal to G 

the distances of the stations ^ a^ 

from the beginning of the line. 

At the points G^, a^, h\ &c., 

erect perpendiculars, G'G, a'a^ 

&c., and make them equal to 

the heights of the respective stations. Through the tops of 

these perpendiculars draw the curved line GK, and it will be 

the profile of the hill in the direction of the line GK. 

On setting out Railway Curves. 

198. It is of course desirable that the line of a railway 
should be perfectly straight and horizontal. This, however, is 
seldom possible for any great distance ; and when it becomes 
necessary to change the direction of the line, it should be done 
gradually by a curve. The curve almost universally employed 
for this purpose is the arc of a circle, and such an arc may be 
traced upon the ground by either of the following methods. 

First Method. — When the centre of the circle can be seen 
from every part of the curve. 

Let AB, CD be two straight portions of a road which it is 
desired to connect by an arc of a circle. Set up a theodolite at 
B and another at C, and from each point range a line at right 
angles to the lines AB and CD respectively ; and at the inter- 
section of these lines, E, which w^ill be the centre of the circle, 
erect a signal which can be seen from any point between B 
and C. The stations must be so . b a, az as a^ f 
chosen that BF equals CF; then 

on these lines drive stakes at I 7^*^^^ 

equal distances, a^^ a^., a^, com- 
mencing from the points B and 
C. li r represents the radius 
of the circle, and d the distance 
between the points a^^ a^., a^^ 
&c., then (Art. 191), 

■y/r'-^d''—T 
will be the distance which must 




Surveying. 



139 



be set off from the first point a^, in the direction ^jE^ to obtain 
a point of the circular arc. In like manner, 

will be the distance to be set off from the point a^, in the di- 
rection a^E ; and, generally, 

Vr-{-{ndy — r 
will be the distance to be set off at the nth points from B and 
C. For example, let r be one mile, or 5280 feet, and d equal 
to 100 feet; then, 

V5280' + 100"^ - 6280 = .94 feet, 
will be the distance aj)^ In a similar manner, we find at 

a^, or 200 feet from B, the offset will be 3.T9 feet. 

^3, or 300 " 

a„ or 400 " " 

a,, or 500 " " 

199. Seco7id Method.— When the 
centre of the circle cannot be seen 
from every part of the curve, the 
offsets may be set off perpendicular- 
ly to the tangent BF, in which case 
they must be computed from the 
formula 



8.52 " 
15.13 " 
23.62 " 



For, in the annexed figure, 



EH=.VGE^-GH^; 
that is, EH=V?^=^^ 




And afi = 


BH r. BE - HE = 7^- Vr' - d\ 




If ^ = 5280 feet, 


we shall find the offsets at intervals 


of 100 


feet to be 


a^^ 


= .95 feet. 






«2^2- 


^ 3.79 " 






^■3^3 


= 8.53 " 






^4^4 


= 15.17 " 






^5^5 


= 23.73 " 




For small distances, the 


offsets will be given w^ith sufiicient 


accuracy by the fo 


rmula 








2?' 


see Art. 191. 





140 Tkigonometey. 

It is very common for surveyors, after they have found the 
first point, 5^ of the curve, to join the points B, &j, and produce 
the line Bh^, to the distance d, and from the end of this line 
set off an otfset to determine the point h^ ; then, producing the 
line Z>j(^2' ^^^ ^^ ^ third offset to determine the point ^3, and 
so on. The objection to this method is, that any error com- 
mitted in setting out one of the points of the curve will occa- 
sion an error in every succeeding one. Whenever this method, 
therefore, is employed, it should be checked by determining 
the position of every fourth or fifth point by independent 
computation and measurement. 

200. Tliird MetJiod. — Where the radius of the curve is 
small, place a theodolite at B, and point its telescope toward 




C Place another theodolite at C, and point its telescope tow- 
ard E, the point of intersection of the lines AB, CD produced. 
Then, if the former be moved through any number of degrees 
toward <2j, and the latter the same number of degrees toward 
{2j, the point a^ will be a point of the curve, for the angle 
B^,C will be equal to BCD {Georn., Prop. 16, B. III.). In the 
same manner, a^^ a^^ &c., any number of points of the curve, 
may be determined. It will be most convenient to move the 
theodolites each time through an even number of degrees, for 
example, an arc of two degrees, and a stake must be driven at 
each of the points of intersection <:/.,, a^^^ a^^ &c. The accuracy 
of this method is independent of any undulations in the sur- 
face of the ground, so that in a hilly country this method may 
be preferable to any other. 

When the position of one end of the curve is not absolutely 
determined, the engineer may proceed more rapidly. Suppose 
it is required to trace an arc of a circle having a curvature of 
two degrees for a hundred feet. 

Place a theodolite at C, the point where the curve com- 



SlJEVEYING. 141 

mences, and lay off from the line CE, toward B, an angle of 
two degrees, and in tlie direction of the axis of the instrument 




set off a distance of 100 feet, which will give the first point a^ 
of the curve. Next lay off from CE an angle of four degrees, 
and from a^ set off a distance of 100 feet, and the point where 
this line cuts the axis of the instrument produced will be the 
second point a^. In the same manner, lay off from CE an an- 
gle of six degrees, and from a^ set off a distance of 100 feet, 
and the point where it cuts the axis of the instrument produced 
will be the third point a^. All the points a^, a^^ a^^ &c., thus 
determined lie in the circumference of a circle {Geom., Prop. 
15, B. III.). Circles thus drawn are generally made with a 
curvature of one or two degrees, or some convenient fraction 
of a degree, for every hundred feet. This method is very ex- 
tensively practised in the United States. 

Survetjing Harbors. 

201. In surveying a harbor, it is necessary to determine 
the position of the most conspicuous objects, to trace the out- 
line of the shore, and discover the depth of water in the neigh- 
borhood of the channel. A smooth, level piece of ground is 
chosen, on which a base line of considerable length is meas- 
ured, and station staves are fixed at its extremities. We 
also erect station staves on all the prominent points to be sur- 
veyed, forming a series of triangles covering the entire surface 
of the harbor. The angles of these triangles are now measured 
with a theodolite, and their sides computed. After the prin- 
cipal points have been determined, subordinate points may be 
ascertained by the compass or plane table. 

Let the following figure be a map of a harbor to be sur- 
veyed. We select the most favorable position for a base line, 
which is found to be on the right of the harbor, from A to B. 



142 Teigonometky. 

We also erect station flags at the points C, D, E, F, and Gr. 
Having carefully measured the base line AB, we measure the 



three angles of the triangle ABO, which enables us to compute 
the remaining sides. We then measure the three angles of the 
triangle ACD, and by means of the side AC, just computed, 
we are enabled to compute AD nnd CD. We then measure 
the three angles of the triangle CDF, and by means of the side 
CD, just found, we are enabled to compute CF and DF. Pro- 
ceeding in the same manner with the triangles CEF, DFG, we 
are enabled, after measuring the angles, to compute the sides. 

202. Having determined the main points of the harbor, we 
may proceed to a more detailed survey by means of the chain 
and compass. If it is required to trace the shore, HCK, we 
commence at H, and observe the bearings with the compass, 
and measure the distances with the chain. Where the shore 
is undulating, it is most convenient to run a straight line for 
a considerable distance, and at frequent intervals measure off- 
sets to the shore. 

When a great many objects are to be represented upon a 
map, the most convenient instrument is 

The Plane Table. 

203. The plane table is a board about sixteen inches square, 



Steveting. 



143 



designed 



drawing- 



paper, and lias two 




to receive a sheet of 
plates of brass upon op- 
posite sides, confined by 
screws, for stretching and 
retaining the paper upon 
the board. The margin 
of the board is divided to 
360 de<?rees from a cen- 
tre C, in the middle of 
the board, and these are 
subdivided as minutely 
as the size of the table 
will admit. On one side 
of the board there is usu- 
ally a diagonal scale of 
equal parts. A compass -box is sometimes attached, which 
renders the plane table capable of answering the purpose of a 
surveyor's compass. 

The ruler. A, is made of brass, as long as the diagonal of 
the table, and about two inches broad. A perpendicular sight- 
vane, B, B, is fixed to each extremity of the ruler, and the eye 
looking through one of them, the vertical thread in the other 
is made to bisect any required distant object. 

To the under side of the table a centre is attached with a 
ball and socket, or parallel plate screws, like those of the the- 
odolite, by which it can be placed upon a staff-head ; and the 
table may be made horizontal by means of a detached spirit 
level. 

204. To prepare the table for use, it must be covered with 
drawing-paper. Then set up the instrument at one of the 
stations, for example, B (see fig. on p. 142), and fix a needle in 
the table at the point on the paper representing that station, 
and place the edge of the ruler against the needle. Then di- 
rect the sights to the station A, and by the side of the ruler 
draw a line upon the paper to represent the direction of AB. 
Then, with a pair of dividers, take from the scale a certain 
number of equal parts to represent the base, and lay off this 
distance on the base line. Having drawn the base line, move 
the ruler around the needle, direct the sights to any object, 



144 Teigonometky. 

as L, and, keeping it there, draw a line along the edge of the 
ruler. Then direct the sights in the same manner to any other 
objects which are required to be sketched, drawing lines in 
their respective directions, taking care that the table remains 
steady during the operation. 

!N'ow remove tlie instrument to the other extremity of the 
base A, and place the point of the paper corresponding to that 
extremity directly over it. Place the edge of the ruler on the 
base line, and turn the table about till the sights are directed 
to the station B. Then, placing the edge of the ruler against 
the needle, direct the sights in succession to all the objects ob- 
served from the other station, drawing lines from the point 
A in their several directions. The intersections of these lines 
with those drawn from the point B will determine the posi- 
tions of the several objects on the map. 

In this manner the plane table may be employed for filling 
in the details of a map ; setting it up at the most remarkable 
spots, and sketching by the eye what is not necessary should 
be more particularly determined, the paper will gradually be- 
come a representation of the country to be surveyed. 

To determine the Dejpth of Water. 

205. Let signals be established on the principal shoals and 
along the edges of the channel, by erecting poles or anchoring 
buoys, and let their bearings be observed from two stations of 
the survey. Then in each triangle there will be known one 
side and the angles, from which the other sides may be com- 
puted, and their positions thus become known. Then ascer- 
tain the precise depth of water at each of the buoys, and pro- 
ceed in this manner to determine as many points as may be 
thought necessary. 

If an observer is stationed with a theodolite at each extremity 
of the base line, we may dispense with the erection of perma- 
nent marks upon the water. One observer in a boat may make 
a sounding for the depth of water, giving a signal at the same 
instant to two observers at the extremities of the base line. 
The direction of the boat being observed at that instant from 
two stations, the precise place of the boat can be computed. 
In this way soundings may be made with great expedition. 



Surveying. M5 

There is also another method, still more expeditious, which 
may afford results sufficiently precise in some cases. Let a 
boat be rowed uniformly across the harbor from one station to 
another, for example, from D to G (see fig. on p. IttS), and let 
a series of soundings be made as rapidly as possible, and the 
instant of each sounding be recorded. Then, knowing the 
entire length of the line DG, and the time of rowing over it, 
we may find by proportion the approximate position of the 
boat at each sounding. 

If the soundings are made in tide waters, the times of high 
water should be observed, and the time of each sounding be 
recorded, so that the depth of water at higli or low tide may 
be computed. In the maps of the United States Coast Survey 
the soundings are all reduced to low-water mark, and the num- 
ber of feet which the tide rises or falls is noted upon the 
map. 

206. The results of the soundings may be delineated upon 
a map in the same manner as the observations of level on page 
135. We draw lines joining all those points where the depth 
of water is the same, for example, 20 feet. Such a line is 
seen to be an undulating line running in the direction from 
E to G. We draw another line connecting all those points 
where the depth of water is 40 feet. This line runs somewhat 
to the east of the former line, but nearly parallel with it. We 
draw other lines for depths of 60 feet, &c. The lines being 
thus drawn, a mere glaiice at the map will show nearly the 
depth of water at any point of the harbor. 

207. Examjjylesfor Practice. 

Prob. 1. The angle of elevation of a spire I found to be 39° 

27', and going directly from it 225 feet on a horizontal plane, 

I found the angle to be only 24° 38'. What is the height of 

the spire, and the distance from its base to the second station ? 

Ans. Height 233.02 feet, distance 508.18 feet. ' 

Prob. 2. Wishing to know the distance of an inaccessible 
object, I measured a horizontal base-line 1328 feet, and found 
the angles at the ends of this line were 81° 23' and 13° 19'. 
What was the distance of the object from each end of the 
base-line? A7is., 1151.41 feet, and 1670.35 feet. 

K 



146 Teigonometey. 

Prob. 3. Wishing to know the distance between two inac- 
cessible objects, C and D, I measured a base-line, AB, 3784 
feet, and found the angle BAD .= 47° 32^ the angle DAC==39° 
53', the angle ABC = 46° 34^ and the angle CBD = 38° V. 
What is the distance from C to D ? Ans., 3257.36 feet. 

Prob. 4. Suppose a lighthouse built on the top of a rock ; 
the distance between the place of observation and that part of 
the rock which is level with the eje, and directly under the 
building, is 1860 feet; the distance from the top of the rock 
to the place of observation is 2538 feet, and from the top of 
the building 2550 feet. Required the height of the light- 
house. Ans.^ 17 feet 7 inches. 

Prob. 5. At 85 feet distance from the bottom of a tower, 
standing on a horizontal plane, the angle of its elevation was 
found to be 52° 30\ Required the altitude of the tower. 

Ans., llOi feet. 

Prob. 6. At a certain station, the angle of elevation of an 
inaccessible tower was 26° 30' ; but, measuring 225 feet in a 
direct line toward it, the angle was then found to be 51° 30'. 
Required the height of the tower, and its distance from the 
last station. Ans. Height 186 feet, distance 147 feet. 

Prob. 7. To find the distance of an inaccessible castle gate, 
I measured a line of 73 yards, and at each end of it took the 
angle of position of the object and the other end, and found 
the one to be 90°, and the other 61° 45'. Required the dis- 
tance of the castle from each station. 

Ans.^ 135.8 yards, and 154.2 yards. 

Prob. 8. From the top of a tower 143 feet high, by the sea- 
side, I observed that the angle of depression of a boat was 35°. 
What was its distance from the bottom of the tower? 

Ans., 204.22 feet. 

Prob. 9. I wanted to know the distance between two places, 
A and B, but could not meet with any station from whence I 
could see both objects. I measured a line CD = 200 yards; 
from C the object A was visible, and from D the object B was 
visible, at each of which places I set up a pole. I also meas- 
ured rC = 200 yards, and DE==200 yards, and at F and E set 
up poles. I then measured the angle AFC==83°, ACF=54° 



SUEVEYING. 147 

Sr, ACD = 53° 30^, BDC = 156° 25^ BDE^54:° 30^ and BED 
= 88° 30'. Eequired the distance from A to B. 

Ans., 345.5 yards. 

Proh. 10. From the top of a lighthouse, the angle of de- 
pression of a ship at anchor was 3° 38', and at the bottom of 
the lighthouse the angle of depression was 2° 43'. Required 
the horizontal distance of the vessel, and the height of the 
promontory above the level of the sea, the lighthouse being 
85 feet high. 

Ans. Distance 5296.4 feet, height 251.3 feet. 

Proh. 11. An observer, seeing a cloud in the west, measured 
its angle of elevation, and found it to be 64°. A second ob- 
server, situated half a mile due east from the first station, and 
on the same horizontal plane, found its angle of elevation at 
the same moment of time to be only 35°. Required the per- 
pendicular height of the cloud, and its distance from each ob- 
server. 

Ans. Perpendicular height 935.75 yards, distances 1041.1 
and 1631.4 yards. 

Prob. 12. An observer, seeing a balloon in the north, meas- 
ured its angle of elevation, and found it to be 36° 52'. A 
second observer, situated one mile due south from the first 
station, and on the same horizontal plane, found its angle of 
elevation at the same instant to be only 30° 58'. Required 
the perpendicular height of the balloon, and its distance from 
each observer. 

Ans. Perpendicular height 3.003 miles, distances 5.006 and 
5.837 miles. 

Prob. 13. From a window near the bottom of a house which 
seemed to be on a level with the bottom of a steeple, I found 
the angle of elevation of the top of the steeple to be 40° ; then 
from another window, 21 feet directly above the former, the 
like angle was 37° 30'. What was the height and distance of 
the steeple ? 

Ans. Height 245.51 feet, distance 292.59 feet. 

Prob. 14. Wanting to know my distance from an inaccessi- 
ble object, P, on the other side of a river, and having no instru- 
ment for taking angles, but a chain for measuring distances. 



148 



Teigonometey. 



from eacli of two stations, A and B, which were taken at 300 
yards asunder, I measured in a direct line from the object P 
60 yards, viz., AC and BD each equal to 60 yards; also, the 
diagonal AD measured 330 yards, and the diagonal BC 336 
yards. What was the distance of the object P from each sta- 
tion A and B ? 

A71S. AP = 321.Y6 yards, BP=: 300.09 yards. 
Prdb. 15. Having at a certain (unknown) distance taken the 
angle of elevation of a steeple, I advanced 60 yards nearer on 
level ground, and then observed the angle of elevation to be 
the complement of the former. Advancing 20 yards still 
nearer, the angle of elevation now appeared to be just double 
of the first. Required the altitude of the steeple. 

Ans.^ 74.162 yards. 

Prob. 16. In a garrison there are three 
remarkable objects. A, B, C, wliose dis- 
tances from each other are known to be, 
AB 213, AC 424, and BC 262 yards. I 
am desirous of knowing m}^ position and 
distance at a station, P, from wdiich I 
observed the angle APB, 13° 30', and 
the angle CPB, 29° 50^ 

Ans. AP = 605.7122, BP = 429.6814, CP = 524.2365. 

Prob. 17. Supposing the object B to be on the opposite side 
of the line AC (see fig. to Prob. 16), and that the distances of 
the objects w^ere, AB = 8 miles, AC = 12 miles, and BC = 7|- 
miles; also, the angle APB = 19°, and the angle CPB = 25°. 
It is required to find tlie distances AP, BP, and CP. 

Ans., AP = 9.4711 miles, BP = 16.3369 miles, 
CP = 16.8485 miles. 

Prob. 18. In a pentangular field, beginning witli the south 
side, and measuring round toward the east, the first or south 
side was 27.35 chains, the second 31.15 chains, the third 23.70 
chains, the fourth 29.25 chains, and the fifth 22.20 chains; 
also, the diagonal from the first angle to the third was 38.00 
chains, and that from the third to the fifth was 40.10 chains. 
Required the area of the field. Ans., 117 A. 2 R. 39 P. 

Prob. 19. The f olio win o^ are the dimensions of a five-sided 




SlJEVEYIITG. 



149 



field, ABCDE: the side AB = 19.40 chains, and the angle B 
110° 30'; the side BC = 15.55 chains, and the angle C 117° 
45' ; the side CD = 21.25 chains, and the angle D 91° 20' ; and 
the side DE = 27.41 chains. Required the area of the field. 

Ans., 66 ^. 2 ^. 24 P. 
Proh. 20. From a station, H, near the middle of a field, 
ABCDEF, from which I could see all the angles, I measured 
the distances to the several corners, and measured the angles 
formed at H by those distances^ as follows : 



Distances. 


Angles. 


AH, 43.15 chains; 


AHB, 60° 30' 


BH, 29.82 " 


BHC, 47° 40' 


CH, 35.61 " 


CHD, 49° 50' 


DPI, 50.10 " 


DHE, 57° 10' 


EH, 46.18 " 


EHF, 64° 15' 


FH, 36.06 " 


FHA, 80° 35' 


Required the area of the field. 


A?is.,4:12A.lI2.17P, 



CHAPTER Y. 

NAVIGATION. 

208. l^AviGATiON is the art of conducting a ship at sea 
from one port to another. 

There are two methods of determining the situation of a 
vessel at sea. The one consists in finding by astronomical ob- 
servations her latitude and longitude ; the other consists in 
measuring the ship's course, and her progress every day from 
the time of her leaving port, from which her place may be 
computed by trigonometry. The latter method is the one to 
be now considered. 

209. The figure of the earth is nearly that of a sphere, and 
in navigation it is considered perfectly spherical. The eartKs 
axis is the diameter around which it revolves once a day. 
The extremities of this axis are the terrestrial ^^^^5 / one is 
called the north pole, and the other the south pole. 

The equator is a great circle perpendicular to the earth's 
axis. 

Meridians are great circles passing through the poles of 
the earth. Every place on the earth's surface has its own 
meridian. 

210. The longitude of any place is the arc of the equator 
intercepted between the meridian of that place and some as- 
sumed meridian to which all others are referred. In most 
countries of Europe, that has been taken as the standard me- 
ridian which passes through their principal observatory. The 
English reckon longitude from the Observatory of Greenwich ; 
and in the United States, we have usually adhered to the Eng- 
lish custom, though longitude is frequently reckoned from the 
Observatory of Washington. 

Longitude is usually reckoned east and west of the first me- 
ridian, from 0° to 180°. 

The difference of longitude of two places is the arc of the 



Navigation. 



151 



equator included between their meridians. It is equal to the 
difference of their longitudes if they are on the same side of 
the first meridian, and to the sum of their longitudes if on op- 
posite sides. 

211. The latitude of a place is the arc of the meridian 
passing through the place, which is comprehended between 
that place and the equator. 

Latitude is reckoned north and south of the equator, from 
0° to 90°. 

Parallels of latitude are the circumferences of small circles 
parallel to the equator. 

The difference of latitude of two places is the arc of a 
meridian included between the parallels of latitude passing 
through those places. It is equal to the difference of their 
latitudes if they are on the same side of the equator, and to 
the sum of their latitudes if on opposite sides. 

The distance is the length of the line which a vessel de- 
scribes in a given time. 

The departure of two places is the distance of either place 
from the meridian of the other. If the two places are on 
the same parallel, the departure is the distance between the 
places. Otherwise, we divide the distance AB into portions 
A5, he, cd^ &c., so small 
that the curvature of the 
earth may be neglected. 
Through these points 
we draw the meridians 
PZ>, P(?, &c., and the par- 
allels he^ cf &c. Then 
the departure for A.h is 
eh^ for he it is/b; and 
the whole departure from A to B is el)-\-fc-\-gd-{'hl^] that is, 
the sum of the departures corresponding to the small portions 
into which the distance is divided. 

Distance, departure, and difference of latitude are measured 
in nautical miles, one of which is the 60th part of a degree at 
the equator. A nautical mile is nearly one sixth greater than 
an English statute mile. 

The course of a ship is the angle which the ship's path makes 




152 



Teigonometey. 



with the meridian. A ship is said to continue on the same 
course when she cuts every meridian which she crosses at the 
same angle. The path thus described is not a straight line, 
but a curve called a rlnimh-line. 

The course of a ship is given by the mariner's compass. 

212. The mariner's compass consists of a circular piece of 
paper, called a card, attached to a magnetic needle, which is 
balanced on a pin so as to move freely in any direction. Di- 
rectly over the needle, a line is drawn on the card, one end of 
which is marked IST, and the other S. The circumference is 
divided into thirty-two equal parts called rhumbs ox jpoints^ 
each point being subdivided into four equal parts called quar- 
ter points. 

The points of the compass are designated as follows, begin- 
ning at north and go- 
ing east : north, north 
by east, north -north- 
east, north-east by 
north, north-east, and 
so on, as shown in the 
annexed figure. 

The interval be- 
tween two adjacent 
points is 11° 15', 
which is the eighth 
part of a quadrant. 
On the side of the 
compass-box a black 
line is drawn perpen- 
dicular to the horizon, and the compass should be so placed 
that a line drawn from this mark through the centre of the 
card may be parallel to the keel of the ship. The part of the 
card which coincides with this mark will then show the point 
of the compass to which the keel is directed. The compass is 
suspended in its box in such a manner as to maintain a hori- 
zontal position, notwithstanding the motion of the ship. 

The following table shows the number of degrees and min- 
utes corresponding to each point of the compass : 




Navigatioi^. 



153 



North. 


Pts. 1 


Pts. 


South. 


N. by E. 


K by W. 


1 


11° 15' 


1 


S. by E. 


S. by W, 


N.KE. 


N.N.W. 


2 


22° 30' 


2 


S.S.E. 


S.S.W. 


N.E.byK 


N.W. by N. 


3 


33° 45' 


3 


S.E. by S. 


S.W. by S. 


KE. 


N.W. 


i 


45° 0' 


4 


S.E. 


s.w. 


KE. by E. 


KW.byW. 


5 


56° 15' 


5 


S.E. by E. 


S.W. by W. 


E.N.E. 


W.^^.W. 


6 


67° 30' 


6 


E.S.E. 


W.S.W. 


E. by K. 


W. bv N. 


7 


78° 45' 


7 


E. by S. 


W. by S. 


East. 


West. 


8 


90° 0' 


8 


East. 


West. 



213. The ship's rate of sailing is measured by a log-line. 
The log-line is a cord about 300 yards long, wliich is wound 





round a reel, one end being attached to a piece of thin board 
called a log. This board is in the form of a sector of a circle, 
the arc of which is loaded with lead sufficient to give the board 
a vertical position when thrown upon the water. This is de- 
signed to prevent the log from being drawn along after the 
vessel while the line is running off the reel. 

The time is measured by a sand-glass, through which the 
sand passes in half a minute^ or the 120th part of 
an hour. 

The log-line is divided into equal parts called 
knots, each of which is 50 feet, or the 120th part 
of a nautical mile. Now, since a knot has the 
same ratio to a nautical mile that half a minute 
has to an hour, it follows, that if the motion of a 
ship is uniform, she sails as many miles in an hour 
as she does knots in half a minute. If, then, seven knots are 
observed to run off in half a minute, the ship is sailing at the 
rate of seven miles an hour. 




PLANE SAILING. 
214. Plane sailing is the method of calculating a ship's 
place at sea by means of the properties of a plane triangle. 
Tlie particulars which are given or required are four, viz., the 



154 



Trigonometry. 




Tlirongli 



distance, course, difference of latitude, and departure. Of 
these, any two being given, tlie others may be found. 

Let the figure EPQ represent a portion of the earth's sur- 
face, P the pole, and EQ 
the equator. Let AB 
be a rhumb-line, or the 
track described by a ship 
in sailing from A to B 
on a uniform course. 
Let the whole distance 
be divided into portions 
A5, hc^ &c., so small 

that the curvature of the earth may be neglected 
the points of division draw the meridians PZ>, P<?, &c., and the 
parallels eb^fc^ &c. Then, since the course is everywhere the 
same, each of the angles eKb,fhc, &c., is equal to the course. 
The distances A^, hf^ &c., are the differences of latitude of A 
and J, h and c, &c. Also, eb,fc^ &c., are the departures for the 
same distances. Hence the difference of latitude from A to 
B is equal to 

Ae-Vhf+cg-\-dh, 
and the departure is equal to 

eh-^-fc^gd-^-K^. 

Construct the triangle A'WC so that A!h'e' 
shall be equal to Ahe^ ^^'f shall be equal to 
Ijcf^ c'd'g' equal to cdg^ and d'WM equal to 
dWi. Then A'B' represents the distance sail- "^ 
ed, B'A'C the course, A'C the difference of latitude, and B'C 



Iv V! 



e' 



■f'-Vc 



the departure; that is, the distance, dif- 
ference of latitude, and departure are cor- 
rectly represented b}^ the hypothenuse and 
sides of a right-angled triangle, of which 
the angle opposite to the departure is the 
course. Of these four quantities, any two 
being given, the others may be found. 

Plane sailing does not assume the earth's 
surface to be a plane, and does not involve 
any error even in great distances. 



Departure 




Navigation. 155 

Examples. 

1. A ship sails from A^era Crnz IST.E. by IS". 74 miles. Ee- 
quired her departm-e and difference of latitude. 

According to the principles of right-angled triangles, Art. 51, 
Dejparture — distance x sin. course. 
Diff. latitude = distance x cos. course. 
The course is three points, or 33° 45' ; hence we obtain 

Departure =41.11 miles. 
Diff.latitude = 61.53 miles. 

2. A ship sails from Sandy Hook, latitude 40° 28' N., upon 
a course E.S.E., till she makes a departure of 500 miles. What 
distance has she sailed, and at what latitude has she arrived? 

Bj Trigonometry, Art. 51, 

-^. , departure t^./p 7 ^-^ 7 departure 

JJisiance = —7^ . -Ow' latitude^= -~ . 

sin. course *^ tan. course 

Ans. Distance ==641.20 miles. 

Diff. latitude = 207.11 miles, or 3° 27'. 

Hence the latitude at which she has arrived is 37° 1' IS". 

3. The bearing of Sandy Hook from Bermuda is X. 42° 56' 
W., and the difference of latitude 486 miles. Eequired tho 
distance and djparture. 

By Trigonometry, Art. 54. 

De])arture — diff. latitude x tan. course. 

^ . , diif. latitude 

Distance — —^ 

COS. course. 

Ans. Distance =663.8 miles. 

Departure =452.1 miles. 

4. A ship sails from Bermuda, latitude 32° 22' N"., a distance 
of QQ^ miles, upon a course between north and east, until she 
finds her departure 444 miles. What course has she sailed, 
and what is her latitude? 

By Trigonometry, Art. 51, 

„. departure 

iSin. course =—i^— . 

distance 

Diff, latitude — distance x cos. course. 

Ans. Latitude = 40° 38' IST. 

Course =K41°40'E. 



156 Trigonometry. 

5. The distance from Yera Cruz, latitude 19° 12' K,to Pen- 
sacola, latitude 30° 19' N., is 820 miles. Eequired the bear- 
ing and departure. 

Bj Trigonometry, Art. 51, 

^ dif. latitude 

Cos. coxirsez= '" .. ^ , 

distance 

De^a7'ture = distance x sin. course, 

Ans. Bearing ^K 35° 31' E. 

Departure = 176.95 miles. 

6. A ship sails from Sandy Hook upon a course between 
south and east to the parallel of 35°, when her departure was 
300 miles. Required her course and distance. 

By Trigonometry, Art. 51, 

rr, departure 

Ian. course =i ^ 



cliff, latitude' 

-r. . , diif. latitude 

Distance = -^ . 

cos. course 

A?is. Course S. 12° 27' E. 

Distance 111.5 miles. 

TRAVERSE SAILING. 

215. A traverse is the irregular path of a ship when sail- 
ing on different courses. 

The object of traverse sailing is to reduce a traverse to a 
single course, when the distances sailed are so small that the 
curvature of the earth may be neglected. When a ship sails 
on different courses, the difference of latitude is equal to the 
difference between the sum of the northings and the sum of 
the southings; and, neglecting the earth's curvature, the de- 
parture is equal to the difference between the sum of the 
eastings and the sum of the westings. If, then, the difference 
of latitude and the departure for each course be taken from 
the traverse table, and arranged in appropriate columns, the 
difference of latitude for the whole time may be obtained ex- 
actly, and the departure nearly, by addition and subtraction ; 
and the corresponding distance and course may be determined 
as in plane sailing. 



I^AYIGATION. 



157 



Examples. 
1. A ship sails on the following successive courses : 

1. KE. 23 miles. 

2. E.S.E. 45 

3. E. bjK 3tt 

4. ]N"orth 29 

5. K by W. 31 

6. KKE. 17 
Find the course and distance for the whole traverse. 

We form a table as below, entering the courses from the 
table of rhumbs, page 153, and then enter the latitudes and 
departures taken from the traverse table. 

Traverse Tcible, 



iVo. 


Course. 


Distanre. 


N 


s. 


E. 


w. 


1 


X. 45° E. 


23 


16.26 




16.26 




2 


S. 67° 30' E. 


45 




17.22 


41.57 




3 


N. 78° 45' E. 


34 


6.63 




33.35 




4 


North. 


29 


29.00 








5 


K 11° 15' W. 


31 


30.40 






6.05 


6 


N. 22° 30' E. 


17 


15.71 




6.51 





Sum of columns 



98.00 
17.22 



17.22 



97.69 
6.05 



1.05 



Diff. latitude . . . . = 80.78 K Dep.=.91.64 E. 
Hence the course is found by plane sailing N. 48° 36' E., 
and the distance = 122.2 miles. 
The equations are 

departure 



Tan. course: 
Distance ■. 



diff. latitude' 
diff. latitude 



cos. course 
2. A ship leaving Sandy Hook makes the following courses 
and distances : ^ gj;^ 25 miles. 

2. E.S.E. 32 '' 

3. East 17 

4. E. by S. 51 
6. South 45 
6. S. by E. 63 



158 Teigonometky. 

Required her latitude, the distance made, and the direct 
course. Ans. Latitude = 38° V N. 

Distance = 193.7 miles. 

Course =S.40°4rE. 

3. A ship from Pensacola, latitude 30° 19^, sails on the fol- 
lowing successive courses: 

1. South 48 miles. 

2. S.S.W. 23 " 

3. S.W. 32 " 

4. S.W. bj S. 76 " 

5. West 17 " 

6. W.S.W. 54 " 
Required her latitude, direct course, and distance. 

Ans. Latitude = 27° 23' K 
Course =S. 38°39'W. 
Distance = 225.0 miles. 

4. A ship from Bermuda, latitude 32° 22', sails on the fol- 
lowing successive courses : 

1. N.E. 66 miles. 

2. N.N.E. 14 " 

3. KE. by E. 45 " 

4. East 21 " 
6. E.byK 32 " 

Required her latitude, direct course, and distance. 

Ans. Latitude = 33° 53' N. 
Course =N. 57° 22' E. 
Distance = 168.4 miles. 

216. When the water through which a ship is moving has 
a progressive motion, the ship's progress is affected in the same 
manner as if she had sailed in still water, with an additional 
course and distance equal to the direction and motion of the 
current. 

Ex. 5. If a ship sail 125 miles l^.N.E. in a current which 
sets W. by N. 32 miles in the same time, required her true 
course and distance. 

Form a traverse table containing the course sailed by the 
ship and the progress of the current, and find the difference 
of latitude and departure. The resulting course and distance 
is found as in the preceding examples. 





Xayigatiox. 
Traverse Table. 






Courses. 


Distance. X. 


E. 


^. 


Is^. 22° 30' E. 
Is. TS° 45' W. 


125 , 115.49 
32 1 6.21 


47.84 


31.39 


Dm. latitude 


. . . =121.73 


47.84 
31.39 


31.39 



159 



Departure .... 
Hence the course is found by 
the distance = 122.8 miles. 

Ex. 6. A ship sails S. by E. 
miles an hour ; then S. by ^V. 
miles an hour; and during the 
by X. at the rate of two and a 
the direct course and distance. 

Ans. 



. . . =16.45 E. 
plane sailing X. 7° 42' E., and 

for two hours at the rate of 9 

for ^\Q hours at the rate of 8 

whole time a current sets T^. 

half miles an hour. Eequired 

The course is S. 21° 51' W. 
Distance 57.6 miles. 



PARAXLEL SAILIXG. 
217. Parallel sailing is when a ship sails exactly east or 
west, and therefore remains constantly on the same parallel 
of latitude. In this case the departure is equal to the distance 
sailed, and the difference of longitude may be found by the 
following 

Theoeem. 

The difference of longitude is equal to the distance sailed. 
divided hy the cosine of the latitude of the j)arallel. 

Let P be the pole of the earth, C the centre, AB a portion 
of the equator, and DE any parallel of lati- p 
tude ; then will CA be the radius of the 
equator, and ED the radius of the parallel. 

Let DE be the distance sailed by the ship fU---\ -\e 

on the parallel of latitude, then the difference 
of longitude will be measured by AB, the 
arc intercepted on the equator by the merid- 
ians passing through D and E. -A. 

Since AB and DE correspond to the equal angles ACB, 



160 



Tkigonometet. 



DFE, tliej are similar arcs, and are to each other as their 

radii. Hence 

FD : CA : : arc DE : arc AB. 

But FD is the sine of PD, or the cosine of AD, that is, the 

cosine of the latitude, and CA is the radius of the sphere, 

which is taken as unity; hence 

-r^.^ -, ., 7 distance 

DtTT' lonqitiide = j-^n — r- • 

^ ^ COS. latitude 

Cor. Like portions of different parallels of latitude are to 
each other as the cosines of the latitudes. 

The length of a degree of longitude in different parallels 
may be computed by this theorem. A degree of longitude at 
the equator being 60 nautical miles, a degree in latitude 40° 
may be found by the equation 

The required length = 60x cos. 40° = 45.96. 

The following table is computed in the same manner. 

218. Table sJioicing the length of a degree of longitude for 
each degree of latitude. 



Lat. 


Miles. 


Lat. 1 Miles. 


Lat. 


Miles. 


Lat. 


Miles. 


Lat. 


Miles. 


Lat. 


Miles. 


1 


59.99 


16 


57.68 


31 


51.43 


46 


41.68 


61 


29.09 


76 


14.52 


2 


59.96 


17 


57.38 


32 


50.88 


47 


40.92 


62 


28.17 


77 


13.50 


3 


59.92 


18 


57.06 


33 


50.32 


48 


40.15 


63 


27.24 


78 


12.47 


4 


59.85 


19 i 56.73 


34 


49.74 


49 


39.36 


64 


26.30 


79 


11.45 


5 


59.77 


20 i 56.38 


35 


49.15 


50 


38.57 


65 


25.36 


80 


10.42 


6 


59.67 


21 i 56.01 


36 


48.54 


51 


37.76 


66 


24.40 


81 


9.39 


7 


59.55 


22 I 55.63 


37 


47.92 


52 


36.94 


67 


23.44 


82 


8.35 


8 


59.42 


23 1 55.23 


88 


47.28 


53 


36.11 


68 


22.48 


83 


7.31 


9 


59.28 


24 ^ 54.81 


39 


46.63 


54 


35.27 


69 


21.50 


84 


6.27 


10 


59.09 


25 \ 54.38 


40 


45.96 


55 


34.41 


70 


20.52 


85 


5.23 I 


11 


58.90 


26 i 53.93 


41 


45.28 


56 


33.55 


71 


19.53 


86 


4.19 


12 


58.69 


27 1 53.46 


42 


44.59 


57 


32.68 


72 


18.54 


87 


3.14 1 


13 


58.46 


28 ; 52.98 


43 


43.88 


58 


31.80 


73 


17.54 


88 


2.09 


14 


58.22 


29 52.48 


44 


43.16 


59 


30.90 


74 


16.54 


89 


1.05 


15 


57.96 


30 ! 51.96 


45 


42.43 


60 


30.00 


75 


15.53 


90 


0.00 



Let ABC represent a right-angled triangle ; then, by Trig- 
onometry, Art. 50, 

COS. B 
But. by the preceding Theorem, we have 

T./r. -7 distance 

diff.long.= , 

^^'^- ^^<^^- A. Departure 

from which we see that if one leg of a 
right-angled triangle represent the distance run on any paral- 




The difference of longitude = ^, o oc'/ ^§12^3^13° 32'. 



Nayigatioit. 161 

lei, and the adjacent acute angle be made equal to the degrees 
of latitude of that parallel, then the hypotlienuse will repre- 
sent the difference of longitude. 

Examples. 

1. A ship sails from Sandy Hook, latitude 40° 28' N"., lon- 
gitude 74° 1' W., 618 miles due east. Eequired her present 
longitude. 

61^ 
COS. 40° 28^ 

This, subtracted from 74° V, leaves 60° 29' W., the longi- 
tude required. 

2. A ship in latitude 40° sails due east through nine de- 
grees of longitude. Required the distance run. 

A71S., 413.66 miles. 

3. A ship having sailed on a parallel of latitude 261 miles, 
finds her difference of longitude 6° 15'. What is her latitude ? 

A71S. Latitude 45° 54'. 

4. Two ships in latitude 52° I^., distant from each other 95 
miles, sail directly south until their distance is 150 miles. What 
latitude do they arrive at ? A^is. Latitude 13° 34'. 

MIDDLE LATITUDE SAILING. 

219. By the method just explained may be found the lon- 
gitude wliich a ship makes while sailing on a parallel of lati- 
tude. When the course is oblique, the departure yiay be 
found by plane sailing, but a difficulty is found in converting 
this departure into difference of longitude. 

If a ship sail from A to 33, the depart- p^ 
ure is equal to eb+fc + gd+hB, which is 
less than AC, but greater than DB. Nav- 
igators have assumed that the departure 
was equal to the distance between the me- 
ridians PA, PB, measured on a parallel 
EF, equidistant from A and B, called the 
middle latitude. 

The middle latitude is equal to half the 
sum of the two extreme latitudes, if both are north or both 




162 



Teigonometry. 



south ; but to half their difference^ if one is north and the 
other south. 

The principle assumed in middle latitude sailing is not per- 
fectly correct. For long distances the error is considerable, 
but the method is rendered perfectly accurate by applying to 
the middle latitude a correction which is given in the accom- 
panying tables, page 203. 

220. It has been shown that when a ship sails upon an 
oblique course, the distance, departure, and difference of lati- 
tude may be represented by the sides of a right-angled triangle. 
The difference of longitude is derived from the departure, 
in the same manner as in parallel sailing, the ship being sup- 
posed to sail on the middle latitude parallel. Hence, if we 
combine the triangle ABC for plane sailing j) 
with the triangle BCD for parallel sailing, 
we shall obtain a triangle ABD, by which 
all the cases of middle latitude sailing may 
be solved. 

In the triande BCD, 



BD== 



BC 




COS. CBD' 
that is, the difference of longitude is equal 
to the departure divided l>y the cosine of 
middle latitude. 

In the triangle ABD, since the angle D is the complement 
of CBD, which represents the middle latitude, we have 

sin. D : AB : : sin. A : BD ; 
that is, cosine of middle latitude is to the distance, as the sine 
of the course is to the difference of longitude. 
In the triangle ABC, we have the equation 

BC==ACtan.A. 
But we have before had the equation 
BC=BD cos. CBD. 
Hence, BD cos. CBD = AC tan. A ; 

or, COS. CBD : AC : : tan. A : BD ; - 

that is, cosine of middle latitude is to the difference of latitude, 
as the tangent of the course is to the difference of longitude. 

The middle latitude should always be corrected according 
to the table on page 203. The given middle latitude is to be 



Inavigatiox. 163 

looked for either in the first or h^st vertical column, opposite 
to which, and under the given difference of latitude, is inserted 
the proper correction in minutes, which must be added to the 
middle latitude to obtain the latitude in which the meridian 
distance is exactly equal to the departure. Thus, if the mid- 
dle latitude is 41°, and the difference of latitude 14°, the cor- 
rection will be found to be 25', which, added to the middle 
latitude, gives the corrected middle latitude -11° 25'. 

Examples. 

1. Find the bearing and distance of Liverpool, latitude 53° 
22' K, longitude 2° 52' W., from Xew York, latitude 40° -±2' 
jS"., longitude 71° 1' ^Y. 

Here are a'iveu two latitudes and lono;itudes to find the 
course and distance. 

The difference of latitude is . . . . 12° 10'= 760'. 

The difference of longitude is . . . .71° 9'=rl269'. 

The middle latitude is 17° 2'. 

To which add the correction from p. 203 22'. 

The corrected middle latitude is . . . 17° 21'. 

Then, according to the third of the preceding theorems, 
Diff. led. : COS. mid. lat. : : dif. long. : tern. course = S. 75° 16' E. 

To find the distance by plane sailing, 

-rs. _, diif. latitude ^^^. , ., 

DisteLnce=-^ = 29SS.1 miles. 

cos. course 

2. A ship sailed from Bermuda, latitude 32° 22' X., longi- 
tude 61° 3S' YT., a distance of 500 miles, upon a course W.X.YT. 
Eequired her latitude and longitude at that time. 

By plane sailing, 

Diff. leLtitude = distance X cos. course = 191\S. 

Therefore the required latitude is 35° 33' ; 

the middle latitude 33° 58'; 

and the corrected middle latitude 33° 59'. 

Then we have 

Cos. mid. lat. : distance : : sin. course : diff. long. — oh'l'.l. 
Therefore the longitude required is 73° 55'. 

3. A ship sails south-easterly from Sandy Hook, latitude 
10° 2S' X., longitude 71° V W., a distance of 395 miles. 



164 



Trigonometry. 



when her latitude is 34° 40' jST. Eequired her oonrse and lon- 
gitude. Ans. Course S. 28° 14' E. 

Longitude 70° 5' W. 
4. A ship sails from Brest, latitude 48° 23' IST., longitude 4° 
29' W., npon a course W.S.W., till her departure is 556 miles. 
Required the distance sailed and the place of the ship. 

Ans. Distance 601.8 miles. 
Latitude 44° 33' K 
Longitude 17° 57' W. 

MERCATOR'S SAILING. 
221. Mercator's sailing is a method of computing differ- 
ence of longitude on the principles of Mercator's chart. On 
this chart the meridians, instead of converging toward the 
poles as they do on the globe, are dra^wn pa7'allel to each other, 
bj which means the distance of the meridians is everywhere 

60' 30" 0" 30° 60° .go° 120° iSo" 



150 



120 



go 




made too great except at the equator. To compensate for 
this, in order that the outline of countries may not be too 
much distorted, the degrees of latitude are proportionally en- 
larged, so that the distance between the parallels of latitude 
increases from the equator to the poles. In latitude 60° the 
distance of the meridians is twice as great, compared with a 
degree at the equator, as it is upon a globe, and a degree of 
latitude is here represented twice as great as near the equator. 



ISTAYIGATIOIf. 165 

The diameter of an island in latitude 60° is represented twice 
as great as if it was on the equator, and its area four times 
too great. In latitude 70° 32^ the distance of the meridians 
is three times too great, in latitude 75° SV four times too great, 
and so on, by which means the relative dimensions of coun- 
tries in high latitudes is exceedingly distorted. On this ac- 
count it is not common to extend the chart beyond latitude 
75°. 

222. The distance of any parallel upon Mercator's chart 
from the equator has been computed, and is exhibited in the 
accompanying table, pages 196-202, which is called a Table of 
Meridional Parts. This table may be computed in the fol- 
lowing manner : 

According to Art. 218 the difference of longitude is equal 
to the departure divided by the cosine of the latitude ; or the 
departure multiplied by the secant of the latitude; that is, a 
part of the equator, or any meridian = a like part of a parallel 
of latitude x secant of tlie latitude. Hence V of a meridian 
= V of a iKirallel x sec. latitude. 

But on Mercator's chart the distance between the meridians 
is the same in all latitudes ; that is, a minute on a parallel of 
latitude is equal to a minute at the equator, or a geographical 
mile. Hence the length of one minute^ on any jj art of a one- 
ridian^ is equal to the secant of the latitude. Thus, 

T\\Q first minute of the meridian = the secant of V ; 
second " " = '' 2'; 

third " " = " 3'; 

&c., &c. 

The table of meridional parts is formed by adding together 
the minutes thus found. Thus, 
Mer. parts of l'=:sec. 1'; 
Mer. parts of 2' = sec. V + sec. 2' ; 
Mer. parts of 3' = sec. 1^ + sec. 2^-fsec. 3^ ; 
Mer. parts of 4^ = sec. l^ + sec. 2'-|-sec. 3^ + sec. 4^, 
&c., &c., &c. 

Since the secants of small arcs are nearly equal to unity, 
if the meridional parts are only given to one tentlj of a mile, 
we shall have 



166 



Teigonometky. 



The meridional parts of 1^ = 1.0 mile; 

u a u 2' = 2.0 " 

« " '' 3' ==3.0 " 

" " " 4^=: 40 " &c., 

as shown in the table on page 196. 

At 2° 33^ the sum of the small fractions omitted becomes 
greater than half of one tenth, and the meridional parts of 
2° 33' are 153.1 ; that is, the meridional parts exceed by one 
tenth of a mile the minutes of latitude. At 3° 40' the excess 
is two tenths of a mile; at 4° 21' the excess is three tenths; 
and as the latitude increases, the meridional parts increase 
more rapidly, as is seen from the table. 

An arc of Mercator's meridian contained between two par- 
allels of latitude is called meridional difference of latitude. It 
is found by subtracting the meridional parts of the less lati- 
tude from the meridional parts of the greater, if both are north 
or south, or by adding them together if one is north and the 
other south. Thus, 

The lat. of New York is 40° 42' ; meridional parts = 2677.8, 
" New Orleans 29° 57' ; " " 1884.9 

The true di£E. of lat. is . 10° 45' ; mer. diff. of lat. is 792.9 

If one latitude and the meridional difference of latitude be 
given, the true difference of latitude may be found by revers- 
ing this process. Thus, 

The meridional parts for New Orleans . . . =1884.9 

Meridional difference of latitude between New [ __ 
York and New Orleans ) ~ 

Therefore the meridional parts for New York 
and the corresponding latitude from the 
table is 40° 42'. 

223. If we take the figure ABC used 
in plane sailing, Art. 214, and produce 
AC to E, making AE equal to the me- 
ridional difference of latitude, and draw 
ED parallel to CB, meeting AB produced 
in D, then will DE represent the differ- "^ 
ence of longitude corresponding to the 
departure BC. For we have found that 



792.9 
: 2677.8, 

Diff.Loitg, p 




Navigation. 167 

the true difference of latitude is to the meridioual difference 
of latitude as one is to the secant of the latitude ; and bj Art. 
222 the departure is to the difference of longitude in the same 
ratio; hence, 

The true difference of latitude is to the meridional difference 
of latitude, as the departure is to the difference of longitude. 

Also in the triangle ADE we have 

DE = AExtan.A; 
that is, the difference of longitude is equal to the meridional 
difference of latitude midtiplied ly the tangent of the course. 

ExAilPLES. 

1. Find the bearing and distance from Sandy Hook, latitude 
40° 28' K, longitude 74° V W., to Havre, latitude 49° 29' K, 
longitude 0° 6' E. 

The true difference of latitude is 9° 1'= 641' ; 
meridional difference of latitude = 767.1 ; 

difference of longitude is 74° 7' = 4447'. 

Hence, to£nd the course by the preceding equation, 

tan. course = '^'ff'^^^^' ^^. §0° 13' E. 
mer. diff. tat. 

To find the distance by plane sailing, 

distance — '^ '- = 3183.8 miles. 

COS. course 

2. Find the bearing and distance from Nantucket Shoals, 
in latitude 41° 4' K, longitude 69° 55' W., to Cape Clear, in 
latitude 51° 26' K, longitude 9° 29' TT. 

Ans. Course K 76° E. 

Distance 2572.9 miles. 

3. A ship sails from Sandy Hook a distance of 600 miles 
upon a course S. by E. Required the place of the ship. 

The difference of latitude may be found by plane sailing, 
the difference of longitude by Mercator's sailing. 

Ans. Latitude 30° 39'.5 K ' 
Longitude 71° 36'.7 W. 

4. A ship sails from St. Augustine, latitude 29° 52' N., lon- 
gitude 81° 25' W., upon a course N.E. by E., until her latitude 



168 Tkigonometky. 

is found to be 34° 40^ IsT. What is then her longitude, and 
what distance has she run ? 

Ans. Longitude = 72° 55' W. 
Distance =518.4 miles. 

5. A ship sails from Bermuda upon a course 'N.W. by W. 
until her longitude is found to be 69° 80' W. What is then 
her latitude, and what distance has she run ? 

Ans. Latitude 35° 4' K 
Distance 291.6 miles. 

6. A ship sailing from Madeira, latitude 32° 38' K, longi- 
tude 16° 55' W., steers westerly until her latitude is 40° 2' JST., 
and her departure 2425 miles. Required her course, distance, 
and longitude. Ans. Course K 79° 37' W. 

Distance 2465.3 miles. 
Longitude 67° 9'.3 W. 

7. Find the bearing and distance from Sandy Hook, latitude 
40° 28' F., longitude 74^ 1' W., to the Cape of Good Hope, 
latitude 34° 22' S., longitude 18° 80' E. 

Ans. Course 
Distance 

CHARTS. 

224. The charts commonly used in navigation are pla?ie 
charts, or Mercator^s chart. In the construction of tlie former, 
the portion of the earth's surface whicli is represented is sup- 
posed to be a^lane. The meridians are drawn parallel to each 
other, and the lines of latitude at equal distances. The dis- 
tance between the parallels should be to the distance between 
the meridians, as unity to the cosine of the middle latitude 
of the chart. A chart of moderate extent constructed in this 
manner will be tolerably correct. The distance of the merid- 
ians in the middle of the chart will be exact, but on each side 
it will be either too great or too small. 

When large portions of the earth's surface are to be repre- 
sented, the error of the plane chart becomes excessive. To 
obviate this inconvenience Mercator's chart has been con- 
structed. Upon this chart the meridians are represented by 
parallel lines, and the distance between the parallels of latitude 



ISTavigatioit. 169 

is proportioned to the meridional difference of latitude, as rep- 
resented on page 164. 

We have seen that the tangent of the course is equal to the 
difference of longitude divided by the meridional difference 
of latitude. Hence, while the course remains unchanged, the 
ratio of the meridional difference of latitude to the difference 
of longitude is constant ^ and, therefore, every rhumb line will 
be represented on Mercator's chart by a straight line. This 
property renders Mercator's chart peculiarly convenient to 
navigators. 

The preceding sketch affords a very incomplete view of the 
present state of the science of navigation. The most accurate 
method of ascertaining the situation of a vessel at sea is by 
means of astronomical observations. For these, however, the 
student must be referred to some treatise on Astronomy. 

225. Examjples for Practice. 

Prol), 1. From a ship at sea I observed a point of land to 
bear east by south, and, after sailing north-east 12 miles, I ob- 
served again, and found its bearing to be south-east by east. 
How far was the last observation made from the point of 
land? J.^i., 26.07 miles. 

Prob. 2. If a ship in latitude 50° E"., sails 52 miles in the 
direction south-west by south, what latitude has she arrived 
in, and how much farther to the west ? 

Ans. Latitude 49° 16'.8 K ; west, 28.9 miles. 

Prol). 3. Two ships sail from the same port ; the one sails 
east-northeast 85 miles, the other sails east by south till the 
first ship bears north-west by west. What is the distance of 
the second ship from the port, and also from the first ship ? 
Ans. From the port, 184.7 miles ; from the first ship, 
123.4 miles.' 

Prob. 4. Two ports lie east and west of each other; a ship 
sails from each, namelj^, the ship from the west port sails north- 
east 89 leagues, and the other sails 80 leagues, when she meets 
the former. Kequired the latter ship's course, and the distance 
between the two ports. 

Ans. Course, K 38° 8' W. ; distance, 112.3 leagues. 



170 Tkigonometey. 

Prob. 6. Two ships sail from a certain port ; the one sails 
south by east 45 leagues, and the other south-southwest 64 
leagues. What is the bearing and distance of the first ship 
from the second ? 

Ans. Bearing, K 65° 44' E. ; distance, 36.5 leagues. 

Prob. 6. A ship sailing north-west, two islands appear in 
sight, of which the one bears north, and tlie other west-north- 
west ; but, after sailing 20 leagues, the former bears north- 
east, and the latter west by south. What is the distance 
asunder of the two islands ? Ans.^ 32.38 leagues. 

Prob. 7. To a vessel sailing on a certain course, a headland 
was observed to bear due west ; four hours after which it w^as 
seen at west-southwest ; and six hours after tliis, the vessel 
continuing to run at the same rate, its bearing was found to be 
south-southwest. What was the vessel's course at the time ? 
Solution. ^n..,N.42°S6'W. 

Denote the angle ACB by x. 
Then sin. 22|-° : sin. a^ : : 4 : AB ; 

sin. 45° : sin. (67i° + a^) : : 6 : AB. 

Hence, 

2 sin. 45° . . .„^,o , 

3^ij^722T^sm.aj=sm.(67i -fa^); 

or, 1.231839 sin. x = &m. (67i-°-f a?) 

= sin. 67^° COS. a?+cos. 67J° sin. x. 
Hence, .8491556 sin. a? = . 9238795 cos. x 

'-^^ tan. a^=: 1.0880. 

COS. X 

Hence, ACB = 47° 25' ; that is, ]SrCBz=42° 35'. 

Prob. 8. Two ships of war, intending to cannonade a fort, 
are, by the shallowness of the water, kept so far from it that 
they suspect their guns cannot reach it with effect. In order, 
therefore, to measure the distance, they separate from each 
other 500 rods ; then each ship observes the angle which the 
other ship and the fort subtend, which angles are 38° 16' and 
37° 9'. What, then, is the distance between each ship and the 
fort? Ans., 312 rods and 320 rods. 

Prob. 9. A ship from the Latitude 42° 18' N., sails south- 
west by south until her latitude is 40° 18' K What direct 




[N'avigation. 171 

distance has she sailed, and how many miles has she sailed to 
the westward ? 

Ans. Distance run 144.3 miles, and has sailed to westward 
80.2 miles. 
Prob. 10. A ship having run due east for three days, at the 
rate of eight knots an hour, finds she has altered her longitude 
15 degrees. What parallel of latitude did she sail on ? 

Ans. Latitude 50^" 12'. 
Prob. 11. A ship in latitude 43° 30' K, and longitude 44° 
W., sails south-easterly 532 miles, until her departure from the 
meridian is 420 miles. Required the course steered, and the 
latitude and longitude of the ship. 

Ans. Course S. 52° 8' E., latitude 38° 3'.5 K, 
longitude 34° 45' W. 
Prob. 12. A ship from latitude 43° 20' K, and longitude 52° 
W., sails E.S.E. until her departure is 745 miles. Required 
the distance sailed, and the latitude and longitude of the ship. 
Ans. Distance 806.4 miles, latitude 38° 11'.5 K, 
longitude 35° 36' W. 
Prob. 13. If the height of the mountain called the Peak of 
Teneriffe be 4 miles, and the angle taken at the top of it, as 
formed between a plumb-line and a line conceived to touch the 
earth in the horizon, or farthest visible point, be 87° 25' 55'', 
it is required from hence to determine the magnitude of the 
whole earth, and the utmost distance that can be seen on its 
surface from the top of the mountain, supposing the earth to 
be a perfect sphere. 

Ans. Distance 178.458 miles, diameter 7957.793 miles. 

Prob. 14. Required the course and distance from St. Jago, 
one of the Cape Yerd islands, in latitude 14° 56' E"., to the 
island of St. Helena, in latitude 15° 45' S., their difference of 
longitude being 30° 12'. 

Ans. Course S. 44° 12' E., distance 2567.8 miles. 

Prob. 15. A ship from the latitude of 49° 57' K, and longi- 
tude of 30° W., sails S. 39° W., till she arrives in the latitude 
of 45° 31' N. Required the distance run, and the longitude 
of the ship. 

Ans. Distance 342.3 miles, longitude of ship 35° 21' W. 



172 Tkigonometey. 

Proh. 16. Find the bearing and distance from San Francisco, 
latitude 37° 48' K, longitude 122° 28' W., to Jeddo, latitude 
35° 40' N., longitude 139° 40' E., by Mercator's sailing, 

Ans. Course S. 88° 26' W., distance 4705 miles. 
Prob. 17. Find the bearing and distance from San Fran- 
cisco to Batavia in Java, latitude 6° 9' S., longitude 106° 53' 
E., by Mercator's sailing. 

Ans, Course S. 70° 12' W., distance 7783 miles. 
Prob. 18. Find the bearing and distance from San Fran- 
cisco to Port Jackson, latitude 33° 51' S., longitude 151° 14' 
E., by Mercator's sailing. 

Ans. Course S. 48° 18' W., distance 6462 miles. 
Prob. 19. Find the bearing and distance from San Fran- 
cisco to Otaheite, latitude 17° 29' S., longitude 149° 29' W., 
by Mercator's sailing. 

Ans. Course S. 24° 44' W., distance 3652 miles. 
Prob. 20. Find the bearing and distance from San Fran- 
cisco to Valparaiso, latitude 33° 2' S., longitude 71° 41' W., by 
Mercator's sailing. 

Ans. Course S. 33° 47' E., distance 5113 miles. 



CHAPTER YI. 

SPHERICAL TRIGONOMETRY. 

226. Spherical trigonometiy teaches how to determine 
the several parts of a spherical triangle from having certain 
parts given. 

A spherical triangle is a portion of the surface of a sphere, 
bounded by three arcs of great circles, each of which is less 
than a semi-circumference. 



RIGHT-ANGLED SPHERICAL TRIANGLES. 
Theorem I. 

227. In any right -angled spherical triangle^ the sine of 
either oblique angle is equal to the quotient of the sine of the 
opposite side divided hy the sine of the hypothenuse. 

Let ABC be a spherical triangle, right-angled at A ; then 
will the sine of the angle ABC be equal to the quotient of the 
sine of AC divided by the sine of BC. O 

Let D be the centre of the sphere ; > 

join AD, BD, CD, and draw CE per- / 

pendicular to DB, which will, there- y 

fore, be the sine of the hypothenuse -q^... ./ 

BC. From the point E draw the ^^^\^^ // ^ j / 
straight line EF, in the plane ABD, E^^-^^^ 

perpendicular to BD, and join CF. ^ 

Then, because DB is perpendicular to the two lines CE, EF, 
it is perpendicular to the plane CEF ; and, consequently, the 
plane CEF is perpendicular to the plane ABD {Geom., Prop. 
6, B. YIL). But the plane CAD is also perpendicular to the 
plane ABD; therefore their line of common section, CF, is 
perpendicular to the plane ABD ; hence CFD, CFE are right 
angles, and CF is the sine of the arc AC. 

Kow, in the right-angled plane triangle CFE, 



174 Teigonometey. 

CF 

sin.CEF-gJ. 

But since CE and FE are both at right angles to DB, the 
angle CEF is equal to the inclination of the planes CBD, 
ABD ; that is, to the spherical angle ABC. Tlierefore, 

K-nr^ sin. AC 

Sm. ABCr=:-^ -7777. 

sm. BC 

228. Cor. 1. In any right-angled spherical triangle, the 
sines of the sides m^e proportional to the sines of the oj)posite 
angles. 

For by the preceding theorem, 

sin. AC . . -T^^ 
- — YTT^ =sin. ABU, 
sm. BC ' 

sin. AB . . ^-p. 
and, - — 577 = sin.ACB. 

' sin. BC 

sin. AC _ sin. ABC 

^^^^^^' sin.AB~sin.ACB* 

Cor. 2. In any right-angled spherical triangle, the cosine of 
the hypothenuse is equal to the yi^oduct of the cosines of the 
two sides. 

Let ABC be a spherical triangle, right-angled at A. De- 
scribe the circle DE, of which B is :f 
the pole, and let it meet the three yx^ 
sides of the triangle ABC produced \ \ 
in D, E, and F. Then, because BD \ )^ 
and BE are quadrants, the arc DF is 
perpendicular to BD. And since 

BAC is a rio^ht an^le, the arc AF is ^ . .^ 

perpendicular to BD. Hence the 
point F, where the arcs FD, FA in- 
tersect each other, is the pole of the arc BD {Geom.^ Prop. 5, 
Cor. 2, B. IX.), and the arcs FA, FD are quadrants. 

Now, in the triangle CEF, right-angled at the point E, ac- 
cording to the preceding theorem, we have 

sin. CFE = ^!"'^^ ; that is, sin. CE = sin. CF x sin. CFE. 
sm. CF' ' 

But CF is the complement of AC, CE is the complement 

of BC, and the angle CFE is measured by the arc AD, which 




Spherical Trigonometry. 



175 



is the complement of AB. Therefore, in the triangle ABC, 

we have 

COS. BC = cos. AC X COS. AB. 

Cor. 3. In any right-angled spherical triangle, the cosine of 
either oUique angle divided ly the cosine of its o;pj)osite side 
is equal to the sine of the other oblique angle. 

For in the triano:le CEF we have 



sin. EF 



,.= sin.ECF. 



sin.CF 

But sin. CF is equal to cos. CA, EF is tlie complement of 

ED, which measures the angle ABC ; that is, sin. EF is equal to 

cos. ABC ; also, sin. ECF is the same as sin. ACB ; therefore, 

cos. ABC 



cos. AC 



si 



in. ACB. 



Theorem II. 

229. In any right-angled spherical triangle, the tangent of 
either ohlique angle is equal to the tangent of the ojp])osite side 
divided hy the sine of the adjacent side. 

Let ABC be a spherical triangle, right-angled at A ; then 

be equal to the tangent of AC di- 
vided by the sine of AB. 

Let D be the centre of the sphere ; 
join AD, BD, CD ; draw AE perpen- 
dicular to BD, which will, therefore, 
be the sine of the arc AB. Also, from 
the point E in the plane BDC, draw 
the straight line EF perpendicular to 
BD, meeting DC produced in F, and 
join AF. Then will AF be perpendicular to the plane ABD, 
because, as was shown in the preceding theorem, it is the com- 
mon section of the two planes ADF, AEF, each perpendicular 
to the plane ADB. Therefore FAD, FAE are right angles, 
and AF is the tangent of the arc AC. 

ISTow, in the triangle AEF, right-angled at A, we have 

tan, AEF .= ^. 
But AE is the sine of the arc AB, AF is the tangent of the 




176 Teigonometey. 

arc AC, and the angle AEF is equal to the inclination of the 
planes CBD, ABD, or to the spherical angle ABC ; hence 

tan. ABC = -. — 7-^7 . 
sin.AB 

230. Cor. 1. In any right-angled spherical triangle, the co- 
sine of the hypothenuse is equal to the product of the cotan- 
gents of the two oblique angles. 

Let ABC be a spherical triangle, right-angled at A. De- 
scribe the circle DEF, o# which B j' 
is the pole, and construct the com- '\\^ 
plemental triangle CEF, as in Cor. 2, \ \ 
Theorem I. \ \^ 

Then, in the triangle CEF, accord- ^1 / \ 

ing to the preceding theorem, we have 

tan. EF 
tan. ECF = -^ — j^. 

sm. CE 

But CE is the complement of BC, -A. 

EF is the complement of ED, the measure of the angle ABC; 
and the angle ECF is equal to ACB, being its vertical angle ; 
hence 

A o-n cot. ABC 

tan. ACB — ^^7^ . 

cos BC 

therefore, cos. BC = - — ..^r-> — cot. ABC x cot. ACB. 
' tan. ACB 

Cor. 2. In any right-angled spherical triangle, the cosine of 
either oblique angle is equal to the tangent of the adjacent side 
multiplied by the cotangent of the hypothenuse. 

For in the complemental triangle CEF, according to the 
preceding theorem, we have 

tan. CE 




tan. CFE = 



Hence in the triangle ABC we have 
cot.AB 



sin.EF* 
3 have 
cot.BC , 
cos. ABC' 



that is, cos. ABC = — '-^r^ — tan. AB x cot. BC. 
' cot. AB 

Napier's Rule of the Circidar Parts. 
231. The two preceding theorems, with their corollaries, 



Spheeical Trigonometet. 177 

are sufficient for the solution of all cases of riglit-angled spher- 
ical triangles, and a rule was invented bj Is'apier by means of 
which these principles are easily retained in mind. 

If, in a right-angled spherical triangle, we set aside the right 
angle, and consider only the five remaining parts of the trian- 
gle, viz., the three sides and the two oblique angles, tlien the 
two sides which contain the right angle, and the complements 
of the other three, viz., of the two angles and the hypothenuse, 
are called the circidar jparts. 

Thus, in the triangle ABC, right-angled at A, the circular 
parts are AB, AC, with the complements of B, BC, and C. 

When, of the five circular parts, any one is taken for the 
middle jpai% then, of the remaining four, the two which are 
immediately adjacent to it on the right and left are called the 
adj acent parts ; and the other two, each of which is separated 
from the middle by an adjacent part, are called opposite parts. 

In every question proposed for solution, three of the circu- 
lar parts are concerned, two of which are given, and one re- 
quired; and of these three, the middle part must be such that 
the other two m.ay be equidistant from it ; that is, may be 
either hoth adjacent or hoth opposite parts. The value of the 
part required may then be found by the following 

Rule of Napiee. 

232. The sine of the middle part is equal to the product of 
the tangents of the adjacent parts^ or to the product of the co- 
sines of the opposite parts. 

It will assist the learner in remembering this rule to remark, 
that the first syllable of each of the words tangent and adja- 
cent contains the same vowel a, and the first syllable of the 
words cosine and opposite contains the same vowel o. 

It is obvious that the cosine of the complement of an angle 
is the sine of that angle, and the tangent of a complement is 
a cotangent, and vice versa. 

In the triangle ABC, if we take the 
side h as the middle part, then the side 
G and the complement of the angle C 
are the adjacent parts, and the com- 
plements of the angle B and of the 

M 




178 Teigonometry. 

hypothennse a are the opposite parts. Tlieiij according to 
Napier's rule, sin. & = tan. c cot. C, 

which corresponds with Theorem II. 
Also, by Napier's rule, 

sin. 5 = sin. a sin. B, 
which corresponds with Theorem I. 

Making each of the circular parts in succession the middle 
part, we obtain the ten following equations : 

sin. 5 = sin. a sin. B=itan. c cot. C. 

sin. c=sin. a sin. C = tan. 1) cot. B. 

cos. B = cos. 5 sin. C — cot. cj^ tan. <?. 

cos. <x=:cos. h cos. c =cot. B cot. C. 

COS. C = cos. G sin. B = cot. a tan. 5. 

233. In order to determine whether the quantity sought 
is less or greater than 90°, the algebraic sign of each term 
should be preserved whenever one of them is negative. If the 
quantity sought is determined by means of its cosine, tangent, 
or cotangent, the algebraic sign of the result will show whether 
this quantity is less or greater than 90° ; for the cosines, tan- 
gents, and cotangents are positive in the first quadrant, and 
negative in the second. But since the sines are positive in 
both the first and second quadrants, when a quantity is deter- 
mined by means of its sine, this rule will leave it ambiguous 
whether the quantity is less or greater than 90°. The am- 
biguity may, however, generally be removed by the following 
rule. 

In every right-angled spherical triangle, an oblique angle 
and its opj)osite side are always of the same species / that is, 
both are greater, or both less than 90°. 

This follows from the equation 

sin. & = tan. c cot. C; 
wherOj since sin. h is always positive, tan. c must always have 
the same sign as cot. C ; that is, the side c and the opposite 
angle C both belong to the same quadrant. 

234. When the given parts are a side and its opposite an- 
gle, the problem admits of two solutions ; for two right-angled 
spherical triangles may always be found, having a side and its 
opposite angle the same in both, but of which the remaining 
sides and the remaining angle of the one are the supplements 




Spherical Tkigonometry. 1T9 

of the remaining sides and the remaining angle of the other. 
Thus, let BCD, BAD be the halves of two great circles, and 
let the arc CA be drawn perpendicu- 
lar to BD; then ABC, ADC are two 
right-angled triangles, having the 
side AC common, and the opposite 
angle B equal to the angle D ; but the side DC is the supple- 
ment of BC, AD is the supplement of AB, and the angle ACD 
is the supplement of ACB. 

Examples. 
1. In the right-angled spherical triangle ABC, there are 
given a = 63° 56' and ^ = 40°. Eequired the other side c, and 
the angles B and C. ^ 

To find the side c. 

Here the circular parts concerned are 
the two legs and the complement of the 
hjpothenuse; and it is evident that if 
the complement of a be made the mid- 
dle part^ h and c will be opposite parts ; hence, by Napier's 
rule, COS. <^==cos. b cos. c ; 

Hence, cos. c= ?^; •.• ^=.54° 59' 49''. 

COS. 

To find the angle B. 

Here h is the middle part^ and the complements of B and a 
are opposite parts ; hence 

sin. 5 = COS. (com p. a)xcos. (comp. B) = sin. a sin. B. 

Hence, sin.B = ^^; •.• B = 45° 41' 25". 

' sm. a 

B is known to be an acute angle, because its opposite side is 
less than 90°. 

To find the angle C. 

Here the complement of C is the middle part ; also 5 and 
the complement of a are adjacent parts ; hence 

cos. C = cot. a tan. h. 
Hence, C = 65° 45' 57". 

Ex. 2. In a right-angled triangle ABC; there are given the 




180 Tkigonometet. 

lijpotlienuse a =91° 42', and the angle B=95° Q\ Eequired 
the remaining parts. 

To find the angle C. 

Make the complement of the hypothenuse the middle part ; 
then COS. a=cot. B cot. C. 

Whence C = 71° 36^ 47^ 

To find the side c. 

Make the complement of the angle B the middle part; and 
we have cos. B=: cot. a tan. c. 

Whence c=71° 32' W. 

To find the side K 

Make the side h the middle part ; then 
sin. J = sin. a sin. B. 
Whence J=95° 22' 30'^ 

h is known to be greater than a quadrant, because its opposite 
angle is obtuse. 

Ex. 3. In the right-angled triangle ABC, the side h is 26° 
4', and its opposite angle- B 36°. Required the remaining 
parts. ( ^ = 48° 22^ 52'^ or 131° 37' 8'^ 

Ans. \ c = 42° 19' 17'', or 137° 40' 43". 
( C = 64° 14' 26", or 115° 45' 34". 
This example, it will be seen, admits of two solutions, con- 
formably to Art. 234. 

Ex. 4. In the right-angled spherical triangle ABC, there are 
given the side c, 54° 30', and its adjacent angle B, 44° 50'. 
Eequired the remaining parts. r C = 65° 49' 53". 

A7is.\ « = 63°10' 4". 
( 5 = 38° 59' 11". 
Why is not the result ambiguous in this case? 

Ex. 5. In the right-angled spherical triangle ABC, the side 
b is 55° 28', and the side c 63° 15'. Eequired the remaining 
parts. ( a = 75° 13' 2". 

^n5.]B = 58°25'47". 
(C = 67°27' 1'^ ■ 

Ex. 6, In the right-angled spherical triangle ABC, there are 



Spherical Teigoxometkt. 



ISl 



given the augle B = 6r 20', and the angle C = 5S° 16^ Ee- 



qnired the remaining parts. ( a — 76° 30^ 37". 

A7is.\ 5==65°28'5S^ 

235. A triangle, in which one of the sides is equal to a 
quadrant, may be solved upon the same principles as right- 
angled triangles, for its polar triangle will contain a right 
angle. See Geom., Prop. 9, B. IX. 

Ex. 7. In the spherical triangle ABC, the side BC=:90°, the 
angle C=:42° 10', and the angle A = 115° 20'. Required the 
remaining parts. 

Taking the supplements of the given parts, we shall have 
in the polar triangle the hvpothenuse a' ==180° -115° 20' = 6^° 
40', and one of the sides, c' = lS0°-42° 10' = 137° 50', from 
which, by Xapier's rule, we find 

B' = 115°.23'20". 
C'=:132° 2' 13". 
^' = 125° 15' 36". 
Hence, taking the supplements of these arcs, we find the 
parts of the required triangle are 

AC = 64° 36' 40". 
AB=:47° 57' 47". 
B = 54°44'24". 
Ex. 8. In the spherical triangle ABC, the side AC:=90°, 
the angle C = 69° 13' 46", and the angle A=:72° 12' 4". Re- 
quired the remaining parts. (AB = 70° 8' 39". 

A7isJ BC = 73° 17' 29". 
( B = 96°13'23". 



OBLIQUE-^VKGLED SPHERICxVL TRIAXGLES. 
Theokem hi. 
236. I?i any spherical triangle, the sines of the sides are 
jyrojportional to the sines of the opijosite angles. 

In the case of right-angled spherical C 

triangles, this proposition has already 
been demonstrated. Let, then, ABC 
be an oblique-angled triangle ; we are 
to prove that 
sin. BC : sin. AC : : sin. A : sin. B. ^ ^B 




182 Tkigonometey. 

Througli the point C draw an arc of a great circle CD per- 
pendicular to AB. Then, in the spherical triangle ACD, 
right-angled at D, we have, by Napier's rule, 
sin. CD = sin. AC sin. A. 
Also, in the triangle BCD, we have 

sin. CD = sin. BC sin. B. 
Hence, sin. AC sin. A=:sin. BC sin. B, 

or, sin. BC : sin. AC : : sin. A : sin. B. 

237. Cor. 1. In any spherical triangle, the cosines of the 
sides are proportional to the cosines of the segments of the 
hase, made hy a perpendicular from the opposite angle. 

For, by Theorem L, Cor. 2, 

^^. cos. AC cos. BC 

cos. CD — -r^^ — f^Y^ . 

cos. AD COS. BD 
Hence cos. AC : cos. BC : : cos. AD : cos. BD. 

Cor. 2. The cosines of the angles at the hase are proportional 
to the sines of the segments of the nertical angle. 

For, by Theorem L, Cor. 3, 

^'''- ^-^ " slrTACD "" siiTBCD * 
Hence cos. A : cos. B : : sin. ACD : sin. BCD. 

Cor. 3. The sines of the segments of the lase are reciprocally 
proportional to the tangents of the angles at the hase. 

For, by Theorem II., 

tan. CD = sin. AD x tan. A = sin. BDxtan. B. 
Hence sin. AD : sin. BD : : tan. B : tan. A. 

Cor. 4. The cotangents of the two sides are proportional to 
the cosines of the segments of the vertical angle. 

For, by Theorem IL, Cor. 2, 

^-^ cos. ACD COS. BCD 
tan. CD = — ^ . ^ — - — — T57T-- 
cot. AC cot. BC 

Hence cos. ACD : cos. BCD : : cot. AC : cot. BC. 

Theorem IY. 

238. If from an angle of a spherical triangle a perpen- 
dicular he drawn to the hase, then the tangent of half the sum 
of the segments of the hase is to the tangent of half the sum 
of the sides, as the tangent of half the difference of the sides 



Spheeical Teigonometey. 



183 




is to the tangent of half the difference of the seg^nents of the 
hase. 

Let ABC be any spherical trian- 
gle, and let CD be drawn from C 
perpendicular to the base AB ; then 
tan. i(BD + AD) : tan. i(BC + AC) : : 
tan.i(BC-AC) : tan. i(BD-AD). 

Let BC==a, AC = l, BD^m, and ^' 
AD=^. Then, by Theorem III., Cor. 1, 

COS. a : COS. h : : cos. m : cos. n. 

Whence, Geora., Prop. 8, B. IL, 
COS. 5 + COS. a : cos. h — cos. a : : cos. n + cos. m : cos. oi — cos. m 

But by Trigonometr}^, Art. 85, 

cos. 5 + cos. a : cos. 5 — cos. a : ; cot. i{a + l) : tan. ^{a—h). 

Also, by the same Art., 
COS. ^ + cos. m : cos. ?i— cos. m : : cot. J-(m + n) : tan. ^{in—n) 

Therefore 
cot. ^{a-{-V) : cot. ^{m-\-n) : : tan. ^{a — lj) : tan. ^{711—71). 

But, since tangents are reciprocally as their cotangents. Art. 
35, we have 

cot. i{a-\-'b) : cot. ^{771 -\-n) : : tan. ^{m-\-n) : tan. ^{a + l). 

Hence 

tan. ^{771 -\- 71) : tan. ^ia-^-V) : : tan. ^{a—l) : tan. ^{m — n). 

239. In the solution of obliqne angled spherical triangles, 
six cases may occur, viz. : 

1. Given two sides and an angle opposite one of them. 

2. Given two angles and a side opposite one of them. 

3. Given two sides and the included angle. 

4. Given two ano-les and the included side. 

5. Given the three sides. 

6. Given the three angles 



Case I. 
240. Given two sides and a7i angle opposite one of them^ 
to find the remaining parts. 

In the triangle ABC, let there be given the two sides AC 
and BC, and the angle A opposite one of them. The angle 
B may be found by Theorem III. 

sin. BC : sin. AC : : sin. A : sin. B. 



184: Tkigonometey. 

From the angle C let fall the perpendicular CD upon the 
side AB. The triangle ABC is di- c 

vided into two right-angled triangles, 
in each of which there is given the 
hjpothenuse and the angle at the 
base. The remaining parts may 
then be found by Napier's rule. ^ 

Ex. 1. In the oblique-angled spherical triangle ABC, the 
side AC=r70° 10' 30'',BC = 80° h' 4.", and the angle A^ZZ"" 
15' 7''. Required the other parts. 

sin. BC : sin. AC : : sin. A : sin. B^31° 34' 38". 

Then, in the triangle ACD, 

COS. AC = cot. A cot. ACD. 

Whence ACD == 77° 27' 47". 

Also, in the triangle BCD, 

COS. BC^rcot.Bcot.'BCD. 
Whence BCD=: 83° 57' 29". 

Therefore ACB = 161° 25' 16". 

To find the side AB. 

sin. A : sin. ACB : : sin. BC : sin. AB^145° 5' 0". 

When we have given two sides and an opposite angle, there 
are, in general, two solutions, each of which will satisfy the 
conditions of the problem. If the side AC, the angle A, and 
the side opposite this angle are given, c 

then, with the latter for radius, de- /4\\ 

scribe an arc cutting tlie arc AB in y^ / \ \ 

the points B and B'. The arcs CB, ^^^^^---^ / j \ 
CB' will be equal, and each of the tri- A^^^^-,^ L^-^r^^'^'^ 
angles ACB, ACB' will satisfy the ^' 

conditions of the problem. There is the same ambiguity in 
the numerical computation. The angle B is found by means 
of its si7ie. But this may be the sine either of ABC, or of its 
sujpjplement AB'C (Art. 34). In the preceding example, the 
first proportion leaves it ambiguous whether the angle B is 
31° 34' 38", or its supplement 148° 25' 22". In order to avoid 
false solutions, we should remember that the greater side of 
a sj)herical triangle must lie ojpjposite the greater angle, and 
conmrsely {Geom,, Prop. 16, B. IX.). Thus, since in the pre- 



Spheeical Teigonometky. 185 

ceding example the side AC is less than BC, the angle B must 
be less than A, and, therefore, cannot be obtuse. 

If the quantity sought is determined by means of its cosine, 
tangent, or cotangent, the algebraic sign of the result will 
show whether this quantity is less or greater than 90° ; for 
the cosines, tangents, and cotangents are positive in the first 
quadrant, and negative in the second. Hence the algebraic 
sign of each term of a proportion should be preserved when- 
ever one of them is negative. 

Ex. 2. In the spherical triangle ABC, the side a=124:° 53^, 
h = 31° 19', and the angle A = 16° 26'. Eequired the remain- 
ing parts. [B=: 10° 19' 34'^ 

Ans.\ C = 171°48'22". 
( c=155° 35' 22". 

Case II. 
241. Given tiuo angles and a side ojpjyosite one of them^ to 
find the remaining jparts. 

In the triangle ABC let there be given two angles, as A and 
B, and the side AC opposite to one ^ 

of them. The side BC may be 
found by Theorem III. > 

sin. B : sin. A : : sin. AC : sin. BC. / 

a/ 

From the unknown angle C draw 
CD perpendicular to AB ; then will "^ 

the triangle ABC be divided into two ris^ht-ano^led trianHes, 
in each of which there is given the hypothenuse and the angle 
at the base. Whence we may proceed by I^apier's rule, as in 
Case I. 

Ex. 1. In the oblique-angled spherical triangle ABC, there 
are given the angle A = 52° 20', B = 63° 40', and the side 
5 = 83° 25'. Required the remaining parts. 

sin. B : sin. A : : sin. AC : sin, BC = 61° 19' 53". 

Then, in the triangle ACD, 

tan. AD =:-^^^-.- AD = 79° 18' 17'^ 
cot. AC 

Also, in the triangle BCD, 



186 



Tkigonometry. 



tan. BD 



COS. B 



.•BD = 39°3'8' 



Hence 



~ cot. BC 

AB = 118'' 2r W 




To find the angle ACB. 
sin. BC : sin. AB : : sin. A : sin. ACB = 127° 26' 47^ 

When we have given two angles and an opposite side, there 
are, in general, two solutions, each of which will satisfy the 
conditions of the problem. If the angle A, tlie side AC, and 
the angle opposite this side are given, then through the point 
C there may generally be 
drawn two arcs of great cir- 
cles CB, CB', making the 
same angle with AB, and 
each of the triangles ABC, 
AB'C will satisfy the condi- 
tions of the problem. There is the same ambiguity in the 
numerical computation, since the side BC is found by means 
of its sine (Art. 34). In the preceding example, however, 
there is no ambiguity, because the angle A is less than B, 
and, therefore, the side a must be less than J, that is, less than 
a quadrant. 

Ex. 2. In the oblique-angled spherical triangle ABC, the 
angle A is 128° 45^ the angle C = 30° 35', and B=:i68° 50'. 
Required the remaining parts. 

It will be observed that in this case the perpendicular BD, 
drawn from the angle B, falls without the triangle ABC, and 
therefore the side AC is the difference between the segments 
CD and AD. ( AB = 37° 28' 20". 

^ AC = 40° 9' 4". 
B =32° 37' 58". 



Ans. 



Case III. 
242. Given two sides and the included angle, to find the re- 
maining parts. 

In the triangle ABC let there be given two sides, as AB, 
AC, and the included angle A. Let fall the pe^rpendicular 
CD on the side AB ; then, by Napier's rule, 
cos. A = tan. AD cot. AC. 



Spheeical Teigonometet. 



18Y 



Having found the segment AD, the segment BD becomes 
known; then, bj Theorem III., Cor. 3, 
sin. BD : sin. AD : : tan. A : tan. B. 

The remaining parts may now be 
found by Theorem III. 

Ex. 1. In the spherical triangle 
ABC, the side AB = 73° 20^ AC=:41° 
45', and the angle A = 30° 30'. Ke- 
qnired the remaining parts, 
cos. A 




tan. AD 



AD = 37° 33' 4r. 



cot. AC 
Hence BD = 35° 46' 19". 

sin. BD : sin. AD : : tan. A : tan. B = 31° 33' 43". 
Also, by Theorem III., Cor. 1, 

COS. AD : cos. BD : : cos. AC : cos. BC = 40° 12' 59". 
Then, by Theorem HI., 

sin. BC : sin. AB : : sin. A : sin. ACB = 131° 8' 46". 
Ex. 2. In the spherical triangle ABC, the side AB=r78° 
15', AC = 56° 20', and the angle A = 120°. Eeqnired the 
other parts. ( B =48° 57' 29". 

Ans. \C =62° 31' 40". 
(BC = 107° 7' 45". 



Case IY. 

243. Given two angles and the included side, to find the 
remaining i^arts. 

In the triangle ABC let there be given two angles, as A 
and ACB, and the side AC inchided 
between them. From C let fall the 
perpendicular CD on the side AB. 
Then, by Napier's rule, 

COS. AC = cot. A cot. ACD. 

Having found the angle ACD, the 
angle BCD becomes known ; then, by 
Theorem III., Cor. 4, 

cos. ACD : COS. BCD : : cot. AC : cot. BC. 

The remaining parts may now be found by Theorem III. 

Ex. 1. In the spherical triangle ABC, tlie angle A = 32° 10', 




188 



Teigoitometey. 



the angle ACB = 133° 20^ and the side AC=:39° 15^ Ke- 
quired the otlier parts. 



Hence, 



cot. ACD =^ ^-^AA2 .,. ACB = 64:° V 57^ 
cot. A 

BCD = 69° 18^ 3^ 



Then 

COS. ACD : COS. BCD : : cot. AC : cot. BC = 45° 20' 43'^ 

Also, by Theorem III., Cor. 2, 

sin. ACD : sin. BCD : : cos. A : cos. B = 28° 15' 47''. 
Then, by Theorem TIL, 

sin. B : sin. ACB : : sin. AC : sin. AB = 76o 23' 5". 
Ex. 2. In the spherical triangle ABC, the angle A = 125° 
20', the angle C = 48° 30', and the side AC = 83° 13'. Ee- 
quired the remaining parts. ( AB= 56° 39' 9". 

^?2,9.-|bC=114° 30' 24". 
( B =62° 54' 38". 



Case Y: 
244. Given the three sides of a spherical triangle^ to find 
the angles. 

the three sides. 
c 



In the triangle ABC let there be given 




From one of the angles, as C, draw CD 
perpendicular to AB. Then, by The- 
orem lY., tan. lAB : tan. .^AC+BC) : : 
tan. i(AC-BC) : tan. |-(AD-BD). 

Hence AD and BD become known ; 
then, by Napier's rule, 

cos. A = tan. AD cot. AC. 

The other angles may now be easily found. 

It is generally most convenient to let fall the perpendicular 
upon the longest side of the triangle. 

Ex. 1. In the spherical triangle ABC, the side AB = 112° 
25', AC = 60° 20', and BC = 81° 10'. Required the angles, 
tan. 56° l^' : tan. 70° 45' : : tan. 10° 25' : tan. 19° 24' 26". 

Hence AD = 36° 48' 4", and BD = 75° 36' 56". 

Then, cos. A = tan. AD x cot. AC •,• A = 64° 46' 36'^ 

Also, cos. B = tan. BDxcot. BC \' B = 52° 42' 12". 

Then sin. AC : sin. AB : : sin. B : sin. ACB = 122° 11' 6". 



Spherical Tkigonometky. 



189 



Ex. 2. In the spherical triangle ABC, the side AB = 4:0° 35', 
AC = 39° 10', and EC = 71° 15'. Eequired the angles. 

r A = 130° 35' 65". 

Ans.\ B= 30° 25' 34". 

( C= 31° 26' 32". 

Case YI. 
245. Given the three angles of a spherical triangle, to find 
the sides. 

If A, B, C are the angles of the given triangle, and a^ h, c its 
sides, then 1S0°-A, 180°-B, and\80°-C are the sides of its 
polar triangle, whose angles may be found by Case Y. Then 
the supplements of those angles will be the sides a, 5, c of the 
proposed triangle. 

Ex. 1. In the spherical triangle ABC, the angle A = 125° 
34', E = 98° 44', and C = 61° 53'. Eequired the sides. 
The sides of the polar triangle are 

54° 26', 81° 16', and 118° T. 
From which, by Case Y., the angles are fonnd to be 

134° 6' 21", 41° 28' 17", and 53° 34' 47". 
Hence the sides of the proposed triangle are 
AB=45° 53' 39", BC = 138° 31' 43", and AC = 126° 25' 13'^ 
Ex. 2. In the spherical triangle ABC, the angle A = 109° 
55', B = 116° 38', and C = 120° 43'. Eequired the sides. 

( a^ 98° 21' 20'^ 

AnsA 5 = 109° 50' 10". 

( c = 115° 13' 7". 



TRIGONOMETRICAL FORMULAE 

246. Let ABC be any spherical 
trian2:le, and from the ans^le B draw 
the arc BD perpendicular to the base 
AC. Eepresent the sides of the trian- 
gle by <z, 5, <?, and the segment AD by 
x\ then will CD be equal toh^x. 
By Theorem HI., Cor. 1, 
cos. c : cos. a : : cos. x : cos (h—x) 

: : cos. X : cos. h cos. .T-fsin. h 




sm.x 



190 Teigonometky. 

(Trig., Art. 82), formula (4). 

Whence 

COS. a COS. c»=:COS. 1} COS. c COS. a?+sin. b cos. c sin. a?; 

or, dividing each term by cos. a?, and substituting the value of 

sin. a? . . ^ _^- - . 

(Art. 35), we obtam 

cos. x^ ^' 

cos. « = cos. h cos. c+sin. J cos. c tan. x. 

But bj Theorem II., Cor. 2, we have 

COS. A cos. A sin. c , . 
tan. X— ■ — 7 — =■ (Ar't. 35). 

cot. C COS. 6' ^ ^ , 

Hence cos. (2 = cos. J cos. <? + sin. J sin. c cos. A, (1) 

from which all the formulas necessary for the solution of 
spherical triangles may be deduced- 

In a similar manner we obtain 

cos. ^:=cos. a COS. c + sin. a sin. c cos. B, (2) 

COS. (? = cos. a COS. J-f-sin. (Z sin. J cos. C. (3) 

These equations express the following Theorem : 

The cosine of either side of a spherical triangle is equal to 
the product of the cosines of the two other sides plus the product 
of the sines of those sides into the cosine of their included angle. 

247. From equation (1) we obtain, by transposition, 

cos. « — COS. 5 COS. c 



COS. A 



sin. h sin. c 



a formula w^iich furnishes an angle of a triangle when the 
three sides are known. 

If we add 1 to each member of this equation, we shall have 

. cos. a + sin. 5 sin. c— COS. ^ cos. c 

1 + COS. A = -. — ^1—. . 

sin. sin. c 

But, by Art. 91, 1 + cos. A = 2 cos. '|-A. 

And, by Art. 81, formula (2), by transposition, 

sin. h sin. c— cos. h cos. c— —cos. (b-\-c). 

Hence, by substitution, we obtain 

** Sin. sin. c 

_ 2 sin. ■J(«-hZ> + c) sin. ^(b-\-G—a) 
~ sin. h sin. c ' 

by Art. 83, formula (12). 



Spheeical Tkigonometey. 191 

If, then, we put s=^{a-]-h-{-c), that is, half the sum of the 
sides, we shall find 

,os.iA = J?^^J^^M^. (4) 

^ V sin. sm. c ^ 

By subtracting cos. A from 1 instead of adding, we shall 
obtain, in a similar manner, 

^ V sin. sin. o ^ ^ 

Either formula (4) or (5) may be employed to compute the 
angles of a spherical triangle when the three sides are known, 
and this method may be preferred to that of Art. 244. 

Ex. 1. In a spherical triangle there are given a=63° 50^, 
^ = 80° 19', and c=120° 47'. Eequired the three angles. 
I-Iere half the sum of the sides is 132° 28' = 5. 

Also, 5 -6^:::. 68° 38'. 

Using formula (4), we have 

log. sine s, 132° 28' . . 9.867862 

log. sine (s-a), 68° 38' . . 9.969075 
-log. sine h, 80° 19' comp. 0.006232 
-log. sine c, 120° 47' comp. 0.065952 
Sum 19.909121 
log. cos. I-A, 25° 45' 19" 9.954560 

Hence the angle A .= 51° 30' 38". 

The remaining angles may be found by Theorem III., or by 
formulse similar to formula (4), 



^ V sin. a sin. c 



sin. a sin. b 
^Ye thus find the angle B= 59° 16' 46", 
and C = 131°28' 36". 

Ex. 2. In a spherical triangle there are given (2 = 115° 20', 
5 = 57° 30', and c = 82° 28'. Eequired the three angles. 

(A = 126°35' 2". 

Ans.l B= 48° 31' 42". 

C= 61° 43' 58". 



192 Tkigonometky. 

248. By means of the polar triangle, we may convert the 
preceding formulse for angles into formulse for the sides of a 
triangle, since the angles of every triangle are the supplements 
of the sides of its polar triangle. Let, then, a' ^ V ^ g\ A^, B', 
Q' represent the sides and. angles of the polar triangle, and v^e 
shall have 

a = lSO°-A', 5 = 1S0°-B^, (?=180°-C^ 
Therefore sin. |-Ar=sin. (90°— |-«') = cos. ^a' ^ 
cos. |-A=:cos. (90°— ifi^^)r=sin. \a' ^ 
sin. ^» = sin. (180°-BO = sin. B', 
sin. c=sin. (180°-C')==sin. C^ 

Also, if we put S^ = half the sum of the angles of the polar 
triangle, we shall have 

«4.^»4.c:^540°-(A^ + F + CO, 
or 5=270°-S^ 

whence sin. s— —cos. S', 

sin. (s-a) = ^m. [90°-(S^- AO]=:cos. (S^-A^, 

sin.(,s-&)=:cos. (S'-BO, 

sin. (5 — c) = cos. (S^ — C). 

By substituting these values in formula (5), Art. 247, and 
omitting all the accents, since the equations are applicable to 
any triangle, we obtain 

cos .,- >o^>(^-B)cos.(8-C) . 
cos.,^-V sin. B sin. ' ^^^ 

and formula (4) becomes 



sin ., / -cos.Scos.(S-A) 
si»-2^-V sin. B sin. C ' ^^^ 

w^hicli formulse enable us to compute the sides of a triangle 
when the three angles are known ; and this method may be 
preferred to that of Art. 245. 

In a similar manner, by means of the polar triangle, we 
derive from formula (1), Art. 246, the equation 

cos. A = cos. a sin. B sin. C — cos. B cos. C ; (8) 
that is, the cosine of either angle of a spherical triangle is equal 
to the product of the sines of the two other angles into the cosine 
of their included side, 7ninus the product of their cosines. 



Spheeical Teigonometet. 193 

Ex. 1. In a spherical triangle ABC, there are given A = 
130° 30^ B = 30° 50', and C = 32° 5\ Kequired the three 
sides. 

Here half the sum of the angles is 96° 42' 30'' = S. 
Also, S-Ar=_33°4r30", 

S-B= 65° 52' 30", 
S-C= 64° 37' 30". 
Using formula (6), we have 

log. COS. (S-B), 65° 52' 30" . 9.611435 
log. COS. (S - C), 64° 37' 30" . 9.631992 
-log. sin. B, 30° 50' comp. 0.290270 
-log. sin. C, 32° 5' comp. 0.274781 
Sum 19.808478 
log. cos. ^a, 36° 40' 1" 9.904239 

Hence the side 6^=73° 20' 2". 

The remaining sides may be found by Theorem III., or by 
formulas similar to formula (6). 

, /eos.(S-A)cos.(S-C) 

°°"-^ V sin. A sin. C 



'^ V sm. A sm. B 

We thus find the side h = 4:0° 13' 12", 
and c=42° 0' 12". 

Ex. 2. In the spherical triangle ABC, the angle A = 129° 
30', B = 54° 35', and C = 63° 5'. Eequired the three sides. 

(ar=120°57' 5". 
Ans.l 1= 64° 55' 37". 
( c= 82° 19' 0". 
249. Formula (1), Art. 246, will also furnish a new test 
for removing the ambiguity of the solution in Case I. of 
oblique-angled triangles. For we have 

. cos. (2— COS. h cos. c 

COS. A=: : J : . 

sm. sm. c 
ISTow if COS. a is greater than cos. 5, we shall have 
COS. <2>cos. h COS. c, 
or the sign of the second member of the equation will be the 
same as that of cos. a, since the denominator is necessarily 



194 Teigonometky. 

positive, and cos. c is less than unity. Hence cos. A and cos. 
a will have the same sign ; or A and a will be of the same 
species when cos. <^>cos. J, or sin. <^<sin. h ; that is, 

If the sine of the side ojpjposite to the required angle is less 
than the sine of the other given side, there will he hut one tri- 
angle. 

But if cos. a is less than cos. h, then such a value may be 
given to c as to render 

cos. <3J<cos. h COS. <?, 
or the sign of the second member of the equation will depend 
upon the value of cos. c ; that is, c may be taken so as to ren- 
der COS. A either positive or negative. Hence 

If the sine of the side opjposite to the required angle is 
greater than the sine of the other given side, there loill he two 
triangles which fulfil the given conditions. 

250. Formula (8), Art, 248, will furnish a test for remov- 
ing the ambiguity in Case II. of oblique-angled triangles. For 
we have 

COS. A -f COS. B cos. C 

COS. a= — —. — t5— ; — T^j ; 

sin. B sm. C 

from which it follows, as in the preceding article, that if cos. 
A is greater than cos. B, A and « will be of the same species. 
But if cos. A is less than cos. B, then such values may be 
given to C as to render cos. a either positive or negative. 
Hence 

If the sine of the angle opposite to the required side is less 
than the sine of the other given angle, there will he hut one tri- 
angle / 

But, if the sine of the angle opposite to the required side is 
greater than the sine of the other given angle, there will he two 
triangles which fulfil the given conditions. 

SAILING ON" AN AEC OF A GREAT CIRCLE. 

251. It is demonstrated in Geom., Prop. 6, B. IX., that the 
shortest path from one point to another on the surface of a 
sphere is the arc of a great circle which joins the two given 
points. Hence, if it is desired to sail from one port to another 
by the shortest route, it is necessary to follow an arc of a great 



Spheeical Teigonometey. 



195 



circle, and tliis arc generally does not coincide with a rhumb- 
line. 

The bearing and distance from one place to another on the 
arc of a great circle may be computed from the latitudes and 
longitudes of the places by means of Spherical Trigonometry. 

Thus, let P be the pole of the earth, EQ a part of the equa- 
tor, and A and B the two given places comprehended between 
the meridians PE and PQ. Then PA is the complement of 
the latitude of A, PB is the complement of the latitude of B, 
and the angle P is measured by the arc EQ, which is the 
difference of longitude between the two 
places. Hence, in the triangle ABP, we 



have given two sides AP, 



BP, and the in- 



cluded angle P, from which we may com- 
pute the side AB, and the angles A and B, 
according to Case III. of oblique-angled tri- 
angles. 

Ex. 1. Required the course and distance 
from Nantucket Shoals, in latitude 41° 4^ K, 
longitude 69° 55' W., to Cape Clear, in lati- 
tude 51° 26' E"., longitude 9° 29' W., on the arc of a great circle. 

Here we have given 

the ande P = 69° 55'- 9° 29' = 60° 26' : 




the side PA = 90° 
the side PB = 90° 
cos. P 



Then tan.PD 



-41° 4'=48°56'; 
-51° 26' = 3S° 34'. 

• PD = 21° 28' 35". 



cot. PB 

Whence AD =: 27° 27' 25". 

Also sin. AD : sin. PD : ; tan. P : tan. A r=54° 27' 21", 
and sin. A : sin. PB : : sin. P : sin. AB^^ir 47' 28". 

41° 47' 28" is equal to 2507.47 nautical miles. 

Hence the course from Kantucket 
Shoals to Cape Clear is N. 64° 27' E., 
and the distance is 2507.47 miles. 

According to Mercator's sailing, 
the course on a rhumb-line, found on 
page 167, is K 76° E., and the dis- 
tance 2572.9 miles. Hence the dis- 
tance on an arc of a f^reat circle is 65.4 miles less than on a 




196 TEiaONOMETET. 

rhumb-line, and the former course is 21|- degrees more north- 
erly than the latter. 

"While sailing on a rhumb-line the course of a ship remains 
always the same, but while sailing on an arc of a great circle 
the course is continually changing. The preceding course is 
that with which the ship starts from E"antucket, and a new 
computation of the course should be made every day or two ; 
or it might be more convenient to compute beforehand the 
position of the points in which the great circle intersects the 
meridians for every five degrees of longitude, and the ship 
might be steered upon a direct course for these points succes- 
sively. 

Ex. 2. Kecpired the course and distance from ITantucket 
Shoals to Gibraltar, in latitude 36° 6' N., longitude 6° 20' W., 
on. the shortest route. Ans. The course is E". 73° 29' E. 

Distance 2974.1 miles. 

Ex. 3. Bequired the course and distance from Sandy Hook, 
in latitude 40° 28' 'N., longitude 74° V W., to Madeira, in lati- 
tude 32° 28' N., longitude 16° 55' W., on the shortest route. 

A71S. The course is N. 80° 53' E. 
Distance 2744.1 miles. 

Ex. 4. Kequired the course and distance from Sandy Hook 
to St. Jago, in latitude 14° 54' N., longitude 23° 30' W., on 
the shortest route. Ans. The course is S. 74° 46' E. 

Distance 3037.6 miles. 

Ex. 5. Required the course and distance from Sandy Hook 
to the Cape of Good Hope, in latitude 34° 22' S., longitude 
18° 30' E., on the shortest route. 

Ans, The course is S. 63° 48' E. 
Distance 6792 miles. 

252. Examfplesfor Practice. 

Prob. 1. In the right-angled spherical triangle ABC, there 
are given the angle C 23° 27' 42", and the side I 10° 39' 40". 
Required the angle B, and the sides a and c. 

(c? =11° 35' 49". 

Ans. -^ <? = 4° 35' 26". 

B = 66°58' 1". 



Spherical Teigonometet. 197 

Prob. 2. In the spherical triangle ABC, the side BC = 90°, 
the side AB=32° 57' ^" , and the side AC = 66° 32^ Eequired 
the angles. (A = 132° 2' 44^ 

Ans. ^ B= 42° 56' 12''. 
( C= 23° 49' 26". 

Prob. 3. In the right-angled spherical triangle ABC, there 
are given the angle B=:47° 54' 20", and the angle C = 61° 50' 
29'^ Eequired the sides. r ^ = 61° 4' 56". 

Am.\ 5=40° 30' 20'^ 
( c=50°30'30". 

Prob. 4. In the spherical triangle ABC, the side AC = 90°, 
the side AB = 115° 9', and the angle B = 101° 40'. Eequired 
the remaining parts. ( BC = 113° 18' 7". 

AnsAK =115° 54' 46". 
(C =117° 33' 49". 

Prob. 5. In the spherical triangle ABC, the angle A = 130° 
5' 22", the angle C = 36° 45' 28", and the side AC = 44° 13' 
45". Eequired the remaining parts. 

(AB = 51° 6' 12". 

^W5.-^BC=S4°14'29". 

(B =32° 26' 6", 

Prob. 6. In the spherical triangle ABC, the angle A = 33° 
15' 7", B = 31° 34' 38", and C = 161° 25' 17". Eequired the 
sides. (a= 80° 5' 4". 

AnsAb^ 70°10'30'^ 
(c=145° 5' 2". 

Prob. 7. In the spherical triangle ABC, the side AB = 112° 
22' 58", AC = 52° 39' 4", and BC = 89° 16' 53". Eequired 
the angles. (A= 70° 39' 0". 

Ans. \b= 48° 36' 0". 
(C=119°15'0". 

Prob. 8. In the spherical triangle ABC, the side AB = 76° 
35' 36", AC = 50° 10' 30", and the angle A = 34° 15' 3". Ee- 
quired the remaining parts. ( B = 42° 15' 13". 

Ans.\ C =121° 36' 20". 
(bC= 40° O'lO". 

Prob. 9. The latitudes of the observatories of Paris and 



198 Teigonometey. 

Pekin are 48° 50^ W N. and 39° 54^ 13'^ N., and their differ- 
ence of longitude is 114° 7' 30''. What is their distance ? 

A?is. 73° 56' 40". 

Proh. 10. Eequired the course and distance from 'New York, 
latitude 40° 43' K, longitude 74° 0' ^Y., to San Francisco, lat- 
itude 37° 48' K, longitude 122° 28' W., on the shortest route. 
Ans. The course is N. 78° 16' W. 
Distance, 2229.8 nautical miles. 

Proh. 11. Required the course and distance from San Fran- 
cisco, latitude 37° 48' R, longitude 122° 28' W., to Jeddo, in 
latitude 35° 40' K, longitude 139° 40 E., on the shortest route. 
Ans. The course is E". 56° 41' W. 
Distance, 4461.9 nautical miles. 

Proh. 12. Required the course and distance from San Fran- 
cisco to Batavia in Java, latitude 6° 9' S., longitude 106° 53' E., 
on the shortest route. A71S. The course is N. 67° 30' W. 

Distance, 7516 nautical miles. 

Proh. 13. Required the course and distance from San Fran- 
cisco to Port Jackson, latitude 33° 51' S., longitude 151° 14' E.. 
on the shortest route. Ans. The course is S. 60° 17' "W. 

Distance, 6444 nautical miles. 

Prdb. 14. Required the course and distance from San Fran- 
cisco to Otaheite, latitude 17° 29' S., longitude 149° 29' W., on 
the shortest route. A71S. The course is S. 29° 45' W. 

Distance, 3650.3 nautical miles. 

Proh. 15. Required the course and distance from San Fran- 
cisco to Valparaiso, latitude 33° 2' S., longitude 71° 41' W., on 
the shortest route. Ans. The course is S. 40° 41' E. 

Distance, 5108.5 nautical miles. 

Prob. 16. Suppose two ports, one in north latitude 30°, and 
the other in north latitude 40°, the difference of longitude be- 
tween them being 50°. Required the bearing and distance 
from each of those ports to an island that lies in south latitude 
18°, and which is equally distant from both of the said ports. 
Ans, Bearing from first port, S. 40° 52' 9" E. 
Bearing from second port, S. 15° 9' 47" W. 
The distance, 59° 23' 19" = 3563.3 nautical miles. 



Spherical Trigonometky. 



199 



Solution. — Let P be the north pole, 
EQ the equator, A and B the two 
ports, and C the island. From C draw 
CD perpendicular to AB, and CH 
perpendicular to E.Q 

In the triangle PAB there are given 
two sides and the included angle, f i'om 
which we find AB = 
41°35^5r; PABr= 
62° r 6^^ and PBA 
= 87° 49' 28'^ 

In the right - an- 
gled triangle AEF we 

now know one side and the adjacent angle, from 
which we find AEF = 40° 2' 57'', and AE=r50° 
59' 34". Hence ED = 71° 47' 29". 

In the right-angled triangle GDE we now have one side 
and the adjacent angle, from which w^e find DG = 38° 36' 23", 
andDGE = T8°24'4". 

In the right-angled triangle CGII we know one side and 
the opposite angle, from which we find CG = 18° 23' 19". 
Hence CD = 56° 59' 42". 

In the right-angled triangle ACD we know the two legs, 
from which we find AC = 59° 23' 19", and CAD = CBD = 77° 
0' 45". 

Hence, rAC = 40° 52' 9", and QBC = 15° 9' 47", 




THE END. 







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